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Euclid’s “Elements” Redux John Casey Daniel Callahan EUCLID’S “ELEMENTS” REDUX John Casey Daniel Callahan Version 2014-346 “Euclid’s ‘Elements’ Redux” \u00a92012-2014 Daniel Callahan, licensed under the Creative Commons Attribution-ShareAlike 4.0 International License. Selections from the American Heritage\u00ae Dictionary of the English Lan- guage, Fourth Edition, \u00a92000 by Houghton Mifflin Company. Updated in 2009. Published by Houghton Mifflin Company. All rights reserved. Selections from Dictionary.com Unabridged Based on the Random House Dictionary, \u00a9 Random House, Inc. 2013. All rights reserved. Email the editor with questions, comments, corrections, and additions: editor@starrhorse. com Download this book for free at: |http : \/ \/ www . starrhorse . com\/ euclid\/| “Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?” – Paul Halmos “Pure mathematics is, in its way, the poetry of logical ideas.” – Albert Einstein FIGURE 0.0.1. Extreme close-up of a snowflake. (c) 2013 Alexey Kljatov, ALL RIGHTS RESERVED. Used without per- mission. [This image is NOT covered by a Creative Commons license.] Contents IPart 1. Introduction! Chapter 1. About this project I. 1. Contributors & Acknowledgments | II. 2. Dedication! IChapter 2. About Euclid’s “Elements* |2.1. History 12.2. Influence I 12.3. Outline of Elements I 2.4. Euclid’s method and style of presentation Chapter 3. Open Textbooks 3.1. Usage Rights| 3.2. Open Licenses| 3.3. Affordability |3.4. Milestones! 3.5. Instruction! 3.6. Authorship 3.7. Projects| |Chapter 4. Recommended Reading] IPart 2. The “Elements*! Chapter 1. Angles, Parallel Lines, Parallelograms 11.1. Definitions! 11.2. Postulates! 11.3. Axioms! 1.4. Explanation of Terms 1.5. Propositions from Book I: 1-26 1.6. Propositions from Book I: 27-48 Chapter 2. Rectangles CONTENTS 12.1. Definitions! 12.2. Axiomsl |2.3. Propositions from Book II |Chapter 3. Circles 13.1. Definitions! |3.2. Propositions from Book HI |Chapter 4. Inscription and Circumscription” 14.1. Definitions! |4.2. Propositions from Book IV |Chapter 5. Theory of Proportions 15.1. Definitions! |5.2. Propositions from Book V| |Chapter 6. Applications of Proportions 16.1. Definitions! |6.2. Propositions from Book VI |Chapter 7. Infinite Primes 17.1. Definitions! |7.2. The Proposition |Chapter 8. Planes, coplanar lines, and solid angles| 18.1. Definitions! 8.2. Propositions from Book XI: 1-21 |Part 3. Student Answer Key |Chapter 1. Solutions: Angles, Parallel Lines, Parallelograms |Chapter 2. Solutions: Rectangles |Chapter 3. Solutions: Circles |Chapter 4. Solutions: Inscription & Circumscription 105 106 107 123 123 127 184 184 184 209 209 214 221 221 222 284 284 284 285 285 286 304 305 347 351 358 Part 1 Introduction CHAPTER 1 About this project The goal of this textbook is to provide an open, low-cost, readable edition of Euclid’s “Elements” that can be distributed anywhere in the world. (In terms of the American educational system, this textbook may be used in grades 7-12 as well as undergraduate college courses on proof writing). Euclid’s “Elements” was the foremost math textbook in most of the world for about 2,200 years. Many problem solvers throughout history wrestled with Euclid as part of their early education including Copernicus, Kepler, Galileo, Sir Isaac Newton, Abraham Lincoln, Bertrand Russell, and Albert Einstein. However, “The Elements” was abandoned after the explosion of new math- ematics toward the end of the 19th century, including the construction of for- mal logic, a more rigorous approach to proof-writing, and the necessity of al- gebra as a prerequisite to calculus. While the end of the 19th century was the beginning of a mathematical Golden Age (one that we are still in), many considered Euclid to be hopelessly out of date. Should “The Elements” be sufficiently rewritten to conform to the current textbook standards, its importance in geometry, proof writing, and as a case- study in the use of logic may once again be recognized by the worldwide edu- cational community. This is a goal that no one author can accomplish. As such, this edition of Euclid has been released under the Creative Commons Attribution-ShareAlike 4.0 International License. The intent is to take advantage of crowd-sourcing in order to improve this document in as many ways as possible. “Euclid’s ‘Elements’ Redux” began as “The First Six Books of the Elements of Euclid” by John Casey (which can be downloaded from Project Gutenberg: |http : \/\/www . gutenberg . org\/ebooks\/21076| ), the public domain translation of “The Elements” by Sir Thomas L. Heath, information from Wikipedia and other sources with appropriate licensing, and it includes illustrations com- posed on GeoGebra software as well as original writing. The ultimate goal for this document is to contain all 13 books of “The El- ements” (some perhaps in truncated form) and to be translated as many lan- guages as possible. Some may also wish to fork this project in order to rewrite 7 1.1. CONTRIBUTORS & ACKNOWLEDGMENTS 8 Euclid from the ground up, to create a “purist’s” edition (the current edition fa- vors Casey’s amendments to Euclid’s original work), to create a wiki of math- ematics from the ancient world, or for some other reason. Such efforts are welcome. The prerequisites for this textbook include a desire to solve problems and to learn mathematical logic. Some algebra will be helpful, especially propor- tions. This textbook requires its student to work slowly and carefully through each section. The student should check every result stated in the book and not take anything on faith. While this may seem tedious, it is exactly this attention to detail which separates those who understand mathematics from those who do not. The figures in this textbook were created in GeoGebra. They can be found in the images folder in the source files for this textbook. Files with extensions .ggb are GeoGebra files, and files with the .eps extension are graphics files. Instructional videos are also available on YouTube which demonstrate how to use GeoGebra: http:\/\/www.youtube. com\/channel\/UCjrVV46Fijv-Pi5VcFm3dCQ This document was composed using: GeoGebra Linux Mint L Y X Windows 7 Xubuntu Linux http : \/ \/ www . geogebra . org\/ http : \/ \/ www . linuxmint . com\/ http : \/ \/ www . lyx . org\/ http : \/ \/ windows . micr osof t . com\/ http : \/ \/ xubuntu . org\/ Follow me on Twitter: @euclidredux 1.1. Contributors & Acknowledgments \u2022 Daniel Callahan (general editor) \u2022 Deirdre Callahan (cover art using GIMP on Raphael’s “The School of Athens”) \u2022 John Casey (his edition of “The Elements” is this basis for this edi- tion). \u2022 Sir Thomas L. Heath (various proofs) 1.2. DEDICATION 9 Daniel Callahan would like to thank Wally AxmanrQ Elizabeth Behrman^J Karl Eldei0 Thalia Jeffrei-Q Kirk LancasteiQ Phil Parkei^ and Weatherford Colleg^]for time and facilities to work on this project. 1.2. Dedication This book is dedicated to everyone in the educational community who be- lieves that algebra provides a better introduction to mathematics than geome- try http : \/\/www .math . wichita . edu\/~axmann\/ ^http : \/\/webs .wichita. edu\/physics\/behrman\/behr .htm “”http : \/\/karlelder . com ^http : \/\/www .math. wichita. edu\/~jef f res\/ ^ http : \/\/kirk . math .wichita . edu\/ f http : \/\/www .math. wichita. edu\/~pparker\/ http : \/\/www . wc . edu CHAPTER 2 About Euclid’s “Elements” [This chapter has been adapted from an entry in WikipediaQ Euclid’s “Elements” is a mathematical and geometric treatise consisting of 13 books written by the ancient Greek mathematician Euclid in Alexandria c.300 BC. It is a collection of definitions, postulates (axioms), propositions (the- orems and constructions), and mathematical proofs of the propositions. The thirteen books cover Euclidean geometry and the ancient Greek version of ele- mentary number theory. The work also includes an algebraic system that has become known as geometric algebra, which is powerful enough to solve many algebraic problems, including the problem of finding the square root of a num- ber. With the exception of Autolycus’ “On the Moving Sphere”, the Elements is one of the oldest extant Greek mathematical treatises, and it is the oldest extant axiomatic deductive treatment of mathematics. It has proven instru- mental in the development of logic and modern science. The name “Elements” comes from the plural of “element”. According to Proclus, the term was used to describe a theorem that is all-pervading and helps furnishing proofs of many other theorems. The word “element” is in the Greek language the same as “let- ter”: this suggests that theorems in the “Elements” should be seen as standing in the same relation to geometry as letters to language. Later commentators give a slightly different meaning to the term “element”, emphasizing how the propositions have progressed in small steps and continued to build on previous propositions in a well-defined order. Euclid’s “Elements” has been referred to as the most successful and influ- ential textbook ever written. Being first set in type in Venice in 1482, it is one of the very earliest mathematical works to be printed after the invention of the printing press and was estimated by Carl Benjamin Boyer to be second only to the Bible in the number of editions published (the number reaching well over one thousand). For centuries, when the quadrivium was included in the curriculum of all university students, knowledge of at least part of Euclid’s ] http : \/\/en. wikipedia. org\/wiki\/Euclid’ s_Elements Wikipedia\u00ae is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organiza- tion. 10 2.1. HISTORY 11 “Elements” was required of all students. Not until the 20th century, by which time its content was universally taught through other school textbooks, did it cease to be considered something all educated people had read. Imprinted at London by John (Daye, FIGURE 2.0.1. The frontispiece of Sir Henry Billingsley’s first English version of Euclid’s Elements, 1570. 2.1. History 2.1.1. Basis in earlier work. Scholars believe that the Elements is largely a collection of theorems proved by other mathematicians supplemented by 2.1. HISTORY 12 some original work. Proclus, a Greek mathematician who lived several cen- turies after Euclid, wrote in his commentary: “Euclid, who put together the Elements, collecting many of Eudoxus’ theorems, perfecting many of Theaete- tus’, and also bringing to irrefutable demonstration the things which were only somewhat loosely proved by his predecessors”. Pythagoras was probably the source of most of books I and II, Hippocrates of Chios of book III, and Eudoxus book V, while books IV, VI, XI, and XII probably came from other Pythagorean or Athenian mathematicians. Euclid often replaced fallacious proofs with his own more rigorous versions. The use of definitions, postulates, and axioms dated back to Plato. The “Elements” may have been based on an earlier text- book by Hippocrates of Chios, who also may have originated the use of letters to refer to figures. 2.1.2. Transmission of the text. In the fourth century AD, Theon of Alexandria extended an edition of Euclid which was so widely used that it became the only surviving source until Frangois Peyrard’s 1808 discovery at the Vatican of a manuscript not derived from Theon’s. This manuscript, the Heiberg manuscript, is from a Byzantine workshop c. 900 and is the basis of modern editions. Papyrus Oxyrhynchus 29 is a tiny fragment of an even older manuscript, but only contains the statement of one proposition. Although known to Cicero, there is no extant record of the text having been translated into Latin prior to Boethius in the fifth or sixth century. The Arabs received the Elements from the Byzantines in approximately 760; this version, by a pupil of Euclid called Proclo, was translated into Arabic under Harun al Rashid c.800. The Byzantine scholar Arethas commissioned the copying of one of the extant Greek manuscripts of Euclid in the late ninth century. Although known in Byzantium, the “Elements” was lost to Western Europe until c. 1120, when the English monk Adelard of Bath translated it into Latin from an Arabic translation. The first printed edition appeared in 1482 (based on Campanus of Novara’s 1260 edition), and since then it has been translated into many languages and published in about a thousand different editions. Theon’s Greek edition was recovered in 1533. In 1570, John Dee provided a widely respected “Mathemati- cal Preface”, along with copious notes and supplementary material, to the first English edition by Henry Billingsley. Copies of the Greek text still exist, some of which can be found in the Vat- ican Library and the Bodleian Library in Oxford. The manuscripts available are of variable quality and are invariably incomplete. By careful analysis of 2.2. INFLUENCE 13 the translations and originals, hypotheses have been made about the contents of the original text (copies of which are no longer available). Ancient texts which refer to the “Elements” itself and to other mathemat- ical theories that were current at the time it was written are also important in this process. Such analyses are conducted by J.L. Heiberg and Sir Thomas Little Heath in their editions of the text. Also of importance are the scholia, or annotations, to the text. These ad- ditions which often distinguished themselves from the main text (depending on the manuscript) gradually accumulated over time as opinions varied upon what was worthy of explanation or further study. 2.2. Influence The “Elements” is still considered a masterpiece in the application of logic to mathematics. In historical context, it has proven enormously influential in many areas of science. Scientists Nicolaus Copernicus, Johannes Kepler, Galileo Galilei, and Sir Isaac Newton were all influenced by the Elements, and applied their knowledge of it to their work. Mathematicians and philosophers, such as Bertrand Russell, Alfred North Whitehead, and Baruch Spinoza, have attempted to create their own foundational “Elements” for their respective dis- ciplines by adopting the axiomatized deductive structures that Euclid’s work introduced. The austere beauty of Euclidean geometry has been seen by many in west- ern culture as a glimpse of an otherworldly system of perfection and certainty. Abraham Lincoln kept a copy of Euclid in his saddlebag, and studied it late at night by lamplight; he related that he said to himself, “You never can make a lawyer if you do not understand what demonstrate means; and I left my sit- uation in Springfield, went home to my father’s house, and stayed there till I could give any proposition in the six books of Euclid at sight.” Edna St. Vin- cent Millay wrote in her sonnet Euclid Alone Has Looked on Beauty Bare, “O blinding hour, O holy, terrible day, When first the shaft into his vision shone Of light anatomized!” Einstein recalled a copy of the “Elements” and a magnetic compass as two gifts that had a great influence on him as a boy, referring to the Euclid as the “holy little geometry book”. The success of the “Elements” is due primarily to its logical presentation of most of the mathematical knowledge available to Euclid. Much of the ma- terial is not original to him, although many of the proofs are his. However, Euclid’s systematic development of his subject, from a small set of axioms to deep results, and the consistency of his approach throughout the Elements, 2.3. OUTLINE OF ELEMENTS 14 encouraged its use as a textbook for about 2,000 years. The “Elements” still in- fluences modern geometry books. Further, its logical axiomatic approach and rigorous proofs remain the cornerstone of mathematics. FIGURE 2.2.1. The Italian Jesuit Matteo Ricci (left) and the Chinese mathematician Xu Guangqi (right) published the Chi- nese edition of Euclid’s “Elements” in 1607. 2.3. Outline of Elements Books 1 through 4 deal with plane geometry. 2.3. OUTLINE OF ELEMENTS 15 Book 2 is commonly called the “book of geometric algebra” because most of the propositions can be seen as geometric interpretations of algebraic identi- ties, such as a(b + c + …) = ab + ac + … or (2a + b) 2 + 6 2 = 2(a 2 + (a + 6) 2 ). It also contains a method of finding the square root of a given number. Book 3 deals with circles and their properties: inscribed angles, tangents, the power of a point, Thales’ theorem. Book 4 constructs the incircle and circumcircle of a triangle, and constructs regular polygons with 4, 5, 6, and 15 sides. Books 5 through 10 introduce ratios and proportions. Book 5 is a treatise on proportions of magnitudes. Proposition 25 has as a special case the inequality of arithmetic and geometric means. Book 6 applies proportions to geometry: Similar figures. Book 7 deals strictly with elementary number theory: divisibility, prime numbers, Euclid’s algorithm for finding the greatest common divisor, least common multiple. Propositions 30 and 32 together are essentially equivalent to the fundamental theorem of arithmetic stating that every positive integer can be written as a product of primes in an essentially unique way, though Euclid would have had trouble stating it in this modern form as he did not use the product of more than 3 numbers. Book 8 deals with proportions in number theory and geometric sequences. Book 9 applies the results of the preceding two books and gives the infini- tude of prime numbers (proposition 20), the sum of a geometric series (propo- sition 35), and the construction of even perfect numbers (proposition 36). Book 10 attempts to classify incommensurable (in modern language, irra- tional) magnitudes by using the method of exhaustion, a precursor to integra- tion. Books 11 through to 13 deal with spatial geometry: Book 11 generalizes the results of Books 1-6 to space: perpendicularity, parallelism, volumes of parallelepipeds. Book 12 studies volumes of cones, pyramids, and cylinders in detail, and shows for example that the volume of a cone is a third of the volume of the corresponding cylinder. It concludes by showing the volume of a sphere is pro- portional to the cube of its radius by approximating it by a union of many pyramids. Book 13 constructs the five regular Platonic solids inscribed in a sphere, calculates the ratio of their edges to the radius of the sphere, and proves that there are no further regular solids. 2.4. EUCLID’S METHOD AND STYLE OF PRESENTATION 16 2.4. Euclid’s method and style of presentation Euclid’s axiomatic approach and constructive methods were widely influ- ential. As was common in ancient mathematical texts, when a proposition needed proof in several different cases, Euclid often proved only one of them (often the most difficult), leaving the others to the reader. Later editors such as Theon often interpolated their own proofs of these cases. Euclid’s presentation was limited by the mathematical ideas and notations in common currency in his era, and this causes the treatment to seem awk- ward to the modern reader in some places. For example, there was no notion of an angle greater than two right angles, the number 1 was sometimes treated separately from other positive integers, and as multiplication was treated ge- ometrically; in fact, he did not use the product of more than three different numbers. The geometrical treatment of number theory may have been because the alternative would have been the extremely awkward Alexandrian system of numerals. The presentation of each result is given in a stylized form, which, although not invented by Euclid, is recognized as typically classical. It has six different parts: first is the statement of the proposition in general terms (also called the enunciation). Then the setting-out, which gives the figure and denotes partic- ular geometrical objects by letters. Next comes the definition or specification which restates the enunciation in terms of the particular figure. Then the con- struction or machinery follows. It is here that the original figure is extended to forward the proof. The proof itself follows. Finally, the conclusion connects the proof to the enunciation by stating the specific conclusions constructed in the proof in the general terms of the enunciation. No indication is given of the method of reasoning that led to the result, although the data does provide instruction about how to approach the types of problems encountered in the first four books of the Elements. Some scholars have tried to find fault in Euclid’s use of figures in his proofs, accusing him of writing proofs that depended on the specific figures constructed rather than the general underlying logic (especially concerning Proposition II of Book I). However, Euclid’s original proof of this proposition is general, valid, and does not depend on the figure used as an example to illustrate one given configura- tion. 2.4.1. Criticism. While Euclid’s list of axioms in the “Elements” is not ex- haustive, it represents the most important principles. His proofs often invoke axiomatic notions which were not originally presented in his list of axioms. 2.4. EUCLID’S METHOD AND STYLE OF PRESENTATION 17 Later editors have interpolated Euclid’s implicit axiomatic assumptions in the list of formal axioms. For example, in the first construction of Book 1, Euclid uses a premise that was neither postulated nor proved: that two circles with centers at the distance of their radius will intersect in two points. Later, in the fourth construction, he uses superposition (moving the triangles on top of each other) to prove that if two sides and their angles are equal then they are congruent. During these considerations, he uses some properties of superposition, but these properties are not constructs explicitly in the treatise. If superposition is to be consid- ered a valid method of geometric proof, all of geometry would be full of such proofs. For example, propositions 1.1 – 1.3 can be proved trivially by using superposition. Mathematician and historian W. W. Rouse Ball puts these criticisms in perspective, remarking that “the fact that for two thousand years [“The Ele- ments”] was the usual text-book on the subject raises a strong presumption that it is not unsuitable for that purpose.” CHAPTER 3 Open Textbooks [This chapter has been adapted from an entry in WikipediaQ An open textbook is a textbook licensed under an open copyright license and made available online to be freely used by students, teachers and members of the public. Many open textbooks are distributed in other printed, e-book, or audio formats that may be downloaded or purchased at little or no cost. Part of the broader open educational resources movement, open textbooks increasingly are seen as a solution to challenges with traditionally published textbooks, such as access and affordability concerns. Open textbooks were identified in the New Media Consortium’s 2010 Horizon Report as a compo- nent of the rapidly progressing adoption of open content in higher education. 3.1. Usage Rights The defining difference between open textbooks and traditional textbooks is that the copyright permissions on open textbooks allow the public to freely use, adapt, and distribute the material. Open textbooks either reside in the public domain or are released under an open license that grants usage rights to the public so long as the author is attributed. The copyright permissions on open textbooks extend to all members of the public and cannot be rescinded. These permissions include the right to do the following: \u2022 use the textbook freely \u2022 create and distribute copies of the textbook \u2022 adapt the textbook by revising it or combining it with other materials Some open licenses limit these rights to non-commercial use or require that adapted versions be licensed the same as the original. 3.2. Open Licenses Some examples of open licenses are: \u2022 Creative Commons Attribution (CC-BY) https : \/\/en.wikipedia. org\/wiki\/Open_textbook 18 3.3. AFFORDABILITY 19 \u2022 Creative Commons Attribution Share-Alike (CC-BY-SA) \u2022 Creative Commons Attribution Non-Commercial Share-Alike (CC-BY- NC-SA) \u2022 GNU Free Documentation License Waivers of copyright that place materials in the public domain include: \u2022 Creative Commons Public Domain Certification 3.3. Affordability Open textbooks increasingly are seen as an affordable alternative to tra- ditional textbooks in both K-12 and higher education. In both cases, open textbooks offer both dramatic up-front savings and the potential to drive down traditional textbook prices through competition. 3.3.1. Higher Education. Overall, open textbooks have been found by the Student PIRGs to offer 80% or more savings to higher education students over traditional textbook publishers. Research commissioned by the Florida State Legislature found similarly high savings and the state has since imple- mented a system to facilitate adoption of open textbooks. In the Florida legislative report, the governmental panel found after sub- stantial consultation with educators, students, and administrators that “there are compelling academic reasons to use open access textbooks such as: im- proved quality, flexibility and access to resources, interactive and active learn- ing experiences, currency of textbook information, broader professional collab- oration, and the use of teaching and learning technology to enhance educa- tional experiences.” (OATTF, p. i) Similar state-backed initiatives are under- way in Washington, Ohio, California, and Texas. In Canada, the province of British Columbia became the first jurisdiction to have a similar open textbook program. 3.3.2. K-12 Education. Research at Brigham Young University has pro- duced a web-based cost comparison calculator for traditional and open K-12 textbooks. To use the calculator the inputs commercial textbook cost, planned replacement frequency, and number of annual textbook user count are re- quired. A section is provided to input time requirements for adaptation to local needs, annual updating hours, labor rate, and an approximation of pages. The summary section applies an industry standard cost for print-on-demand of the adapted open textbook to provide a cost per student per year for both textbook options. A summed cost differential over the planned period of use is also calculated. 3.4. MILESTONES 20 3.4. Milestones In November 2010, Dr. Anthony Brandt was awarded an “Access to Artistic Excellence” grant from the National Endowment for the Arts for his innova- tive music appreciation course in Connexions. “Sound Reasoning … takes a new approach [to teaching music appreciation]: It presents style-transcendent principles, illustrated by side-by-side examples from both traditional and con- temporary music. The goal is to empower listeners to be able to listen atten- tively and think intelligently about any kind of music, no matter its style. Ev- erything is listening based; no ability to read music is required.” The module being completed with grant funds is entitled “Hearing Harmony”. Dr. Brandt cites choosing the Connexions open content publishing platform because “it was an opportunity to present an innovative approach in an innovative for- mat, with the musical examples interpolated directly into the text.” In December 2010, open textbook publisher Flat World Knowledge was recognized by the American Library Association’s Business Reference and Ser- vices Section (ALA BRASS) by being named to the association’s list of “Out- standing Business Reference Sources: The 2010 Selection of Recent Titles.” The categories of business and economics open textbooks from Flat World Knowl- edge’s catalog were selected for this award and referenced as “an innovative new vehicle for affordable (or free) online access to premier instructional re- sources in business and economics.” Specific criteria used by the American Library Association BRASS when evaluating titles for selection were: A resource compiled specifically to supply information on a certain subject or group of subjects in a form that will facil- itate its ease of use. The works are examined for authority and reputation of the publisher, author, or editor; accuracy; appropriate bibliography; organization, comprehensiveness, and value of the content; currency and unique addition to the field; ease of use for intended purpose; quality and accu- racy of indexing; and quality and usefulness of graphics and illustrations. Each year more electronic reference titles are published, and additional criteria by which these resources are evaluated include search features, stability of content, graphic design quality, and accuracy of links. Works selected are intended to be suitable for medium to large academic and public libraries. 3.5. INSTRUCTION 21 Because authors do not make money form the sale of open textbooks, several organizations have tried to use prizes or grants as financial incentives for writ- ing open textbooks or releasing existing textbooks under open licenses. Con- nexions announced a series of two grants in early 2011 that will allow them to produce a total of 20 open textbooks. The first five titles will be produced over an 18 month time frame for Anatomy & Physiology, Sociology, Biology, Biology for non-majors, and Physics. The second phase will produce an additional 15 titles with subjects that have yet to be determined. It is noted the most expen- sive part of producing an open textbook is image rights clearing. As images are cleared for this project, they will be available for reuse in even more titles. In addition, the Saylor Foundation sponsors an ongoing “Open Textbook Chal- lenge”, offering a $20,000 reward for newly-written open textbooks or existing textbooks released under a CC-BY license. The Text and Academic Author’s Association awarded a 2011 Textbook Ex- cellence Award (“Texty”) to the first open textbook to ever win such recogni- tion this year. A maximum of eight academic titles can earn this award each year. The title “Organizational Behavior” by Talya Bauer and Berrin Erdogan earned one of seven 2011 Textbook Excellence Awards granted. Bauer & Er- dogan’s “Organizational Behavior” open textbook is published by Flat World Knowledge. 3.5. Instruction Open textbooks are flexible in ways that traditional textbooks are not, which gives instructors more freedom to use them in the way that best meets their instructional needs. One common frustration with traditional textbooks is the frequency of new editions, which force the instructor to modify the curriculum to the new book. Any open textbook can be used indefinitely, so instructors need only change editions when they think it is necessary. Many open textbooks are licensed to allow modification. This means that instructors can add, remove or alter the content to better fit a course’s needs. Furthermore, the cost of textbooks can in some cases contribute to the quality of instruction when students are not able to purchase required materials. A Florida governmental panel found after substantial consultation with educa- tors, students, and administrators that “there are compelling academic reasons to use open access textbooks such as: improved quality, flexibility and access to resources, interactive and active learning experiences, currency of textbook information, broader professional collaboration, and the use of teaching and learning technology to enhance educational experiences.” (OATTF, p. i) 3.7. PROJECTS 22 3.6. Authorship Author compensation for open textbooks works differently than traditional textbook publishing. By definition, the author of an open textbook grants the public the right to use the textbook for free, so charging for access is no longer possible. However, numerous models for supporting authors are developing. For example, a start-up open textbook publisher called Flat World Knowledge pays its authors royalties on the sale of print copies and study aids. Other proposed models include grants, institutional support and advertising. 3.7. Projects A number of projects seek to develop, support and promote open textbooks. Two very notable advocates and supporters of open textbook and related open education projects include the William and Flora Hewlett Foundation and the Bill and Melinda Gates Foundation. CHAPTER 4 Recommended Reading 1. “Geometry: Seeing, Doing, Understanding” 3rd edition, by Harold R. Jacobs (ISBN: 978-0716743613). I recommend this title for beginning geome- try students as a primary textbook along with “Euclid’s ‘Elements’ Redux” as a secondary textbook. (An Enhanced Teacher’s Guide and an Improved Test Bank are also available, although both appear to be out of print.) The text- book’s ISBN: 978-0716743613 2. “Book of Proof” 2nd edition, by Richard Hammack. This open textbook is an introduction to the standard methods of proving mathematical theorems. It can be considered a companion volume to any edition of Euclid, especially for those who are learning how to read and write mathematical proofs for the first time. It has been approved by the American Institute of Mathematics’ Open Textbook Initiative and has a number of good reviews at the Mathematical Association of America Math DL and on Amazon. Visit the website at: http : \/ \/ www . people . vcu . edu\/~rhammack\/BookOf Proof \/ index . html 3. Math Open Referenc^J especially the topic of Triangle Center^} 4. Khan Academ^ 5. “The Thirteen Books of Euclid’s Elements”, translation and commen- taries by Sir Thomas Heath in three volumes. Published by Dover Publica- tions, Vol. 1: ISBN 978-0486600888, Vol. 2: ISBN 978-0486600895, Vol. 3: ISBN 978-0486600901. 6. “Euclid’s Elements – All thirteen books in one volume”. Based on Heath’s translation, Green Lion Press, ISBN 978-1888009194. http : \/\/www. mathopenre f . com\/ ^http : \/\/www .mathopenref . com\/trianglecenters . html https : \/\/www . khanacademy . org\/ 23 Part 2 The “Elements” In this textbook, students are expected to construct figures as they are given, step-by-step. This is an essential component to the learning process that cannot be avoided. In fact, this is the impetus behind the historical quote, “There is no royal road to geometry.” That is, no one learns mathematics “for free”. The propositions of Euclid will be referred to as (for example) either Propo- sition 3.32 or [3.32], with chapter and proposition number separated by a pe- riod. Axioms, Definitions, etc., will also be referred to in this way: for example, Definition 12 in chapter 1 will be denoted as [Def. 1.12]. Exercises to problems will be denoted as (for example) [3.5, #1] for exercise 1 of Proposition 3.5. A note on exercises: generally, an exercise is expected to be solved using the propositions, corollaries, and exercises that preceded it. For example, exercise [1.32, #3] should first be attempted using propositions [1.1]-[1.32] as well as all previous exercises. Should this prove too difficult or too frustrating for the student, then he\/she should consider whether propositions [1.33] or later (and their exercises) might help solve the exercise. It is also permissible to use trigonometry, linear algebra, or other contemporary mathematical techniques on challenging problems. CHAPTER 1 Angles, Parallel Lines, Parallelograms The following symbols will be used to denote certain standard geometric shapes or relationships: \u2022 Circles will be denoted by: o \u2022 Triangles by: A \u2022 Parallelograms by: \u25a1 \u2022 Parallel lines by: || \u2022 Perpendicular lines by: _L In addition to these, we shall employ the usual symbols of algebra, +,-,=,, 7^, jt, ^, as well as two additional symbols: \u2022 Composition: 0 For example, suppose we have the segments AB and BC which intersect at the point B. The statement AB + BC refers to the sum of their lengths, but AB 0 BC refers to their composition as one object. See Fig. 1.0.1. FIGURE 1.0.1. Composition: the geometrical object AB 0 BC is a single object composed of two segments, AB and BC. The composition of angles, however, can be written using either + or 0, and in this textbook their composition will be denoted with +. \u2022 Congruence: = Two figures or objects are congruent if they have the same shape and size, or if one has the same shape and size as the mirror image of the other. This means that an object can be reposi- tioned and reflected (but not re-sized) so as to coincide precisely with the other object Q http : \/\/en . wikipedia . org\/wiki\/Congruence_ (geometry) 26 1.1. DEFINITIONS 27 \u2022 Similar: ~ Two figures or objects are similar if they have the same shape but not necessarily the same size. If two similar objects have the same size, they are also congruent. The Point. 1. A point is a zero dimensional object^] A geometrical object which has three dimensions (length, height, and width) is a solid. A geomet- rical object which has two dimensions (length and height) is a surface, and a geometrical object which has one dimension only is a line. Since a point has none of these, it has zero dimensions. The Line. 2. A line is a one dimensional object: it has only length. If it had any height, no matter how small, it would be space of two dimensions. If it had any width, it would be space of three dimensions. Hence, a line has neither height nor width. (This definition conforms to Euclid’s original definition in which a line need not be straight. However, in all modern geometry texts, it is understood that a “line” has no curves. See also [Def 1.4].) 3. The intersections of lines are points. 4. A line without a curve between its endpoints is called a straight line. It is understood throughout this textbook that a line refers exclusively to a straight line. A curved line (such as the circumference of a circle) will never be referred to merely as a line in order to avoid confusion. Lines have no endpoints since they are infinite in length. A line segment or more simply a segment is like a line except that it is finite in length and has two endpoints which occur at its extremities. A ray is like a line in that it is infinite in length; however, it has only one endpoint. See Fig. 1.1.1. Warren Buck, Chi Woo, Giangiacomo Gerla, J. Pahikkala. “point” (version 13). PlanetMath.org. Freely available at http : \/\/planetmath . org\/point 1.1. Definitions 1.1. DEFINITIONS 28 \u00bb* E F * +\u2014 FIGURE 1.1.1. [Def. 1.2, 1.3, 1.4] AB is a line, CD is a seg- ment, and EF is a ray The Plane. 5. A surface has two dimensions, length and height. It has no width; if it had, however small, it would be space of three dimensions. 6. A surface is called a plane whenever two arbitrary points on the surface can be joined by a right angle. 7. Any combination of points, of lines, or of points and lines on a plane is called a plane figure. A plane figure that is bounded by a finite number of straight line segments closing in a loop to form a closed chain or circuit is called a po\/y^o^] 8. Points which lie on the same straight line, ray, or segment are called collinear points. The Angle. 9. The angle made by of two straight lines, segments, or rays extending outward from a common point but in different directions is called a rectilinear angle or simply an angle. 10. The common point of intersection between straight lines, rays, or seg- ments is called the vertex of the angle. 11. A particular angle in a figure will be denoted by the symbol Z and three letters, such as BAC, of which the middle letter, A, is at the vertex. Hence, an angle may be referred to either as ABAC or ZCAB. Occasionally, this notation will be shortened to “the angle at point A” instead of naming the angle as above. See Fig. 1.1.2. 1 http : \/\/en . wikipedia . org\/wiki\/Polygon 1.1. DEFINITIONS 29 FIGURE 1.1.2. [Def 1.11] Notice that both angles could be re- ferred to as ZBAC, ZCAB, or the angle at point A. Also note that A is a vertex. 12. The angle formed by composing two or more angles is called their sum. Thus in Fig. 1.1.3, we have that A ABC \u00a9 ZPQR = ZABR where the segment QP is applied to the segment BC such that the vertex Q falls on the vertex B and the side QR falls on the opposite side of BC from BA. We generally choose to write ZABC + ZPQR = ZABR to express the same concept. Figure 1.1.3. [Def. 1.12] 13. When two segments BA, AD are composed such that BA \u00a9 AD = BD where BD is another segment, the angles ZBAC and ZCAD are called sup- plements of each other (see Fig. 1.1.4). This definition holds when we replace segments by straight lines or rays, mutatis mutandisn : Figure 1.1.4. [Def. 1.13] ^Mutatis mutandis is a Latin phrase meaning “changing [only] those things which need to be changed” or more simply “[only] the necessary changes having been made”. Source: http:\/\/en. wikipedia.org\/wiki\/Mutatis_mutandis 1.1. DEFINITIONS 30 14. When one segment (AC) stands on another (DB) such that the adjacent angles on either side of the first segment are equal (that is, ZD AC = ZCAB), each of the angles is called a right angle, and the segment which stands on the other is described as perpendicular to the other (or it is called the perpendicular to the other). See Fig. 1.1.5. We may also write that AC is perpendicular to DB or more simply that AC JL DB. It follows that the supplementary angle of a right angle is another right angle. Multiple perpendicular lines on a many-sided object may be referred to as the object’s perpendiculars. The above definition holds for straight lines and rays, mutatis mutandis. A line segment within a triangle from a vertex to an opposite side which is also perpendicular to that side is usually referred to an altitude of the triangle, although it could in a general sense be referred to as a perpendicular of the triangle. 15. An acute angle is one which is less than a right angle. ZCAB in Fig. 1.1.4 and ZDAB is Fig. 1.1.6 are acute angles. 16. An obtuse angle is one which is greater than a right angle. ZCAD in Fig. 1.1.4 is an obtuse angle. The supplement of an acute angle is obtuse, and conversely, the supplement of an obtuse angle is acute. 17. When the sum of two angles is a right angle, each is called the comple- ment of the other. 1 Figure 1.1.5. [Def. 1.14] FIGURE 1.1.6. [Def. 1.17] The angle ZBAC is a right angle. Since ZBAC = ZCAD + ZDAB, it follows that the angles ZBAD, ZD AC are each complements of the other. 1.1. DEFINITIONS 31 Concurrent Lines. 18. Three or more straight lines intersecting the same point are called concurrent lines. This definition holds for rays and seg- ments, mutatis mutandis. 19. A system of more than three concurrent lines is called a pencil of lines. The common point through which the rays pass is called the vertex . The Triangle. 20. A triangle is a polygon formed by three segments joined at their endpoints. These three segments are called the sides of the triangle. One side in particular may be referred to as the base of the triangle for explanatory reasons, but there is no fundamental difference between the properties of a base and either of the two remaining sides of a triangle. 21. A triangle whose three sides are unequal in length is called scalene (the left-hand example in Fig. 1.1.7). A triangle with two equal sides is called isosceles (the middle example in Fig. 1.1.7). When all sides are equal, a triangle is called equilateral, (the right-hand example in Fig. 1.1.7). When all angles are equal, a triangle is called equiangular. FIGURE 1.1.7. [Def 3.21] The three types of triangles: scalene, isosceles, equilateral. 22. A right triangle is a triangle in which one of its angles is a right angle, such as the middle example in Fig. 1.1.7. The side which stands opposite the right angle is called the hypotenuse of the triangle. (In the middle example in Fig. 1.1.7, ZEDF is a right angle, so side EF is the hypotenuse of the triangle.) 23. An obtuse triangle is a triangle such that one of its angles obtuse (such as ACABinFig. 1.1.8). 1.1. DEFINITIONS 32 Figure 1.1.8. [Def. 1.23] 24. An acute triangle is a triangle such that each of its angles are acute, such as the left- and right-hand examples in Fig. 1.1.7. 25. An exterior angle of a triangle is one which is formed by extending the side of a triangle. For example, the triangle in Fig. 1.1.8 has had side B A extended to the segment AD which creates the exterior angle ZD AC. Every triangle has six exterior angles. Also, each exterior angle is the supplement of the adjacent interior angle. In Fig. 1.1.8, the exterior angle ZD AC is the supplement of the adjacent interior angle ZCAB. The Polygon. 26. A rectilinear figure bounded by three or more segments is referred to as a polygon. For example, the object in Fig. 1.0.1 is a plane figure but not a polygon. The triangles in Fig. 1.1.7 are both plane figures and polygons. 27. A polygon is said to be convex when it has no re-entrant angle (that is, it does not have an interior angle greater than 180\u00b0). 28. A polygon of four sides is called a quadrilateral. 29. A quadrilateral whose four sides are equal in length is called a lozenge. A lozenge is also a form of rhombu^Jand therefore also a parallelogram. 30. A rhombus which has a right angle is called a square. 31. A polygon which has five sides is called a pentagon) one which has six sides, a hexagon, etcf] “”https : \/ \/en . wikipedia . org\/wiki\/Rhomb us| 6 See also https : \/\/en . wikipedia . org\/wiki\/Polygon 1.1. DEFINITIONS 33 The Circle. 32. A circle is a plane figure formed by a curved line called the circumference such that all segments constructed from a certain point within the figure to the circumference are equal in length. That point is called the center of the circle. FIGURE 1.1.9. [Def. 1.32] oBDA constructed with center C and radius CD. Notice that CA = CB = CD. Also notice that AB is a diameter. 33. A radius of a circle is any segment constructed from the center to the circumference, such as CA, CB, CD in Fig. 1.1.9. Notice that CA = CB = CD. 34. A diameter of a circle is a segment constructed through the center and terminated in both directions by the circumference, such as AB in Fig. 1.1.9. From the definition of a circle, it follows that the path of a movable point in a plane which remains at a constant distance from a fixed point is a circle. Also, any point P in the plane is either inside, outside, or on the circumference of a circle depending on whether its distance from the center is less than, greater than, or equal to the radius. : Figure 1.1.10. [Def. 1.35] 1.2. POSTULATES 34 Other. 35. A segment, line, or ray in any figure which divides the area of a geometric object into two equal halves is called an Axis of Symmetry of the figure (such as AC in the polygon ABCD, Fig. 1.1.10). 36. A segment constructed from any angle of a triangle to the midpoint of the opposite side is called a median of the triangle. Each triangle has three medians which are concurrent. The point of intersection of the three medians is called the centroid of the triangle. \u25a0 FIGURE 1.1.11. [Def. 1.36] Segment CD is a median of A ABC. 37. A locus (plural: loci) is a set of points whose location satisfies or is determined by one or more specified conditions, i.e., 1) every point satisfies a given condition, and 2) every point satisfying it is in that particular locusQ For example, a circle is the locus of a point whose distance from the center is equal to its radius. 1.2. Postulates We assume the following: (1) A straight line, ray, or segment may be constructed from any one point to any other point. Lines, rays, and segments may be subdivided by points into segments or subsegments which are finite in length. (2) A segment may be extended from any length to a longer segment, a ray, or a straight line. (3) A circle may be constructed from any point (which we denote as its center) and from any finite length measured from the center (which we denote as its radius). http : \/\/en.wikipedia. org\/wiki\/Locus_ (mathematics) 1.3. AXIOMS 35 Note: if we have constructed two points A and B on a sheet of paper, and if we construct a segment from A to B, this segment will have some irregularities due to the spread of ink or slight flaws in the paper, both of which introduce some height and width. Hence, it will not be a true geometrical segment no matter how nearly it may appear to be one. This is the reason that Euclid postulates the construction of segments, rays, and straight lines from one point to another (where our choice of paper, application, etc., is irrelevant). For if a segment could be accurately constructed, there would be no need for Euclid to ask us to take such an action for granted. Similar observations apply to the other postulates. It is also worth nothing that Euclid never takes for granted the accomplishment of any task for which a geometrical construction, founded on other problems or on the foregoing postulates, can be provided. 1.3. Axioms Axioms 1-7 and 9 hold for every kind and variety of magnitude. Axioms 8 and 10-12 are strictly geometrical. Note that all Euclidean magnitudes are positive. (1) If we consider three magnitudes such that the first magnitude is equal to the second and the second magnitude is equal to the third, we infer that the first magnitude equals the third. (a) If A = B, and B = C, then A = C. (2) If equals are added to equals, then their sums are equal. (a) If A = B and C is added to both A and B, then A + C = B + C. (3) If equals are taken from equals, then the remainders are equal. (a) If A = B and C is subtracted from both A and B, then A \u2014 C = B-C. (4) If equals are added to unequals, then the sums are unequal. (a) If A > B and C is added to both A and B, then A + C> B + C. (b) If A < B and C is added to both A and B, then A + C B and C is subtracted from both A and B, then A \u2014 C > B-C. (b) If A < B and C is subtracted from both A and B, then A – C < B-C. (6) The doubles of equal magnitudes are equal. (a) If A = B, then 2A = 2B. (7) The halves of equal magnitudes are equal. (a) If A = B, then A\/2 = B\/2. 1.3. AXIOMS 36 (8) Magnitudes which can be made to coincide are equal. (a) The placing of one geometrical object on another, such as a line on a line, a triangle on a triangle, or a circle on a circle, etc., is called superposition. The superposition employed in geometry is only mental; that is, we conceive of one object being placed on the other. And then, if we can prove that the objects coincide, we infer by the present axiom that they are equal in all respects, including magnitude. Superposition involves the following principle which, without being explicitly stated, Euclid makes frequent use: "Any figure may be transferred from one position to another without change in size or form." (9) The whole is equal to the sum of all its parts. This is sometimes stated as: the whole is greater than the sum of its parts. (10) Two straight lines cannot enclose a space. (a) This is equivalent to the statement, "If two straight lines have two points common to both, then they coincide in direction." Al- ternatively, we say that they form a single line because they co- incide at every point. (b) The above holds for segments and rays, mutatis mutandis. (11) All right angles are equal to each other. (a) A proof: Let there be two straight lines AB, CD, and two perpen- diculars to them, namely, EF, GH. Then if AB, CD are made to coincide by superposition, so that the point E will coincide with G, then since a right angle is equal to its supplement, the line EF must coincide with GH. Hence ZAEF = ZCGH. (12) If two straight lines (AB, CD) intersect a third straight line (AC) such that the sum of the two interior angles (ABAC, ZACD) on the same side equals less than two right angles, then if these lines will meet at some finite distance. (This axiom is the converse of [1.17].) See Fig. 1.3.1. 1.4. EXPLANATION OF TERMS 37 FIGURE 1.3.1. [Axiom 1.12] The lines AB and CD must even- tually meet (intersect) at some finite distance. The above holds for rays and segments, mutatis mutandis. 1.4. Explanation of Terms Axioms: "Elements of human reason" are certain general propositions, the truths of which are self-evident, and which are so fundamental that they can- not be inferred from any propositions which are more elementary. In other words, they are incapable of demonstration. "That two sides of a triangle are greater than the third" is, perhaps, self-evident; but it is not an axiom since it can be inferred by demonstration from other propositions. However, we can give no proof of the proposition that "two objects which are equal in length to a third object are also equal in length to each other". Since that statement is self-evident, it is considered an axiom. Propositions which are not axioms are properties of figures obtained by processes of reasoning. They may be divided into theorems and problems. A theorem is the formal statement of a property that may be demonstrated from known propositions. These propositions may themselves be theorems or axioms. A theorem consists of two parts: the hypothesis, or that which is as- sumed, and the conclusion, or that which is asserted to follow from the argu- ment. We present four examples: Theorem. (1) IfX is Y, then Z is W. we have that the hypothesis is that X is Y, and the conclusion is that Z is W. Converse Theorems: Two theorems are said to be converses when the hy- pothesis of either is the conclusion of the other. Thus the converse of the theo- rem (1) is: Theorem. (2) If Z is W, then X is Y. 1.4. EXPLANATION OF TERMS 38 From two theorems (1) and (2), we may infer two others called their con- trapositives. The contrapositive of (1) is: THEOREM. (3) If Z is not W, then X is not Y. The contrapositive of (2) is: THEOREM. (4) IfX is not Y, then Z is not W. Theorem (4) is also called the invers^]of (1), and (3) is the inverse of (2). A problem is a proposition in which something is proposed to be done, such as a line or a figure to be constructed under some given conditions. The solution of a problem is the method of construction which accomplishes the required result. In the case of a theorem, the demonstration is the proof that the conclusion follows from the hypothesis. In the case of a problem, the demonstration is the construction which creates the proposed object. The statement or enunciation of a problem consists of two parts: the data, or that which we assume we have to work with, and that which we must ac- complish. Postulates are the elements of geometrical construction and have the same relation with respect to problems as axioms do to theorems. A corollary is an inference or deduction from a proposition. A lemma is an auxiliary proposition required in the demonstration of a principal proposition. A secant line is a line which cuts (intersects) a system of lines, a circle, or any other geometrical figure. Congruent figures are those that can be made to coincide by superposition. They agree in shape and size but differ in position. Hence by [Axiom 1.8], it follows that corresponding parts or portions of congruent figures are congruent and that congruent figures are equal in every respect. The Rule of Symmetry: If X = Y, it follows that Y = X. "The counterpart of a proposition obtained by exchanging the affirmative for the negative quality of the whole proposition and then negating the predicate: The inverse of "'Every act is predictable' is 'No act is unpredictable.'" The American Heritage\u00ae Dictionary of the English Language, Fourth Edition copyright \u00a92000 by Houghton Mifflin Company Updated in 2009. Published by Houghton Mifflin Company. All rights reserved. 1.5. PROPOSITIONS FROM BOOK I: 1-26 39 1.5. Propositions from Book I: 1-26 Proposition 1.1. CONSTRUCTING AN EQUILATERAL TRIANGLE. Given an arbitrary segment, it is possible to construct an equilateral triangle on that segment. PROOF. We wish to construct an equilateral triangle on the segment AB. With A as the center of a circle and AB as its radius, we construct the circle oBCD [Postulate 1.3]. With B as center and BA as radius, we construct the circle oACE, cutting oBCD at point C. Connect segments CA, CB [Postulate 1.1]. We claim that A ABC is the required equilateral triangle. Figure 1.5.1. [1.1] Because A is the center of the circle oBCD, AC = AB [Def. 1.33]. Since B is the center of the circle oACE, BA = BC. Since AB = BA (i.e., denoting a segment by its endpoints reading from left to right or from right to left does not affect the segment's length), by [Axiom 1.1], we have that AC = AB = BA = BC, or simply AC = AB = BC. Hence, AABC is an equilateral triangle [Def. 1.21]. Since AABC is con- structed on the given segment AB, the proof follows. \u25a1 Examination questions. 1. What is assumed in this proposition? 2. What is that we were to have accomplished? 3. What is a finite straight line? 4. What is the opposite of finite? 5. What postulates were cited and where were they cited? 6. What axioms were cited and where were they cited? 7. What use is made of the definition of a circle? What is a circle? 8. What is an equilateral triangle? Exercises. 1.5. PROPOSITIONS FROM BOOK I: 1-26 40 The following exercises use Fig. 1.5.1 and are to be solved when the student has completed Chapter 1. 1. If the segments AF, BF are joined, prove that the figure [HACBF is a rhombus. 2. If AB is extended to the circumferences of the circles (at points D and E), prove that the triangles ACDF and ACEF are equilateral. 3. If CA, CB are extended to intersect the circumferences at points G and H, prove that the points G, F, H are collinear and that the triangle AGCH is equilateral. 4. Connect CF and prove that CF 2 = 3AB 2 . 5. Construct a circle in the space ACB bounded by the segment AB and the partial circumferences of the two circles. Proposition 1.2. CONSTRUCTING A STRAIGHT-LINE SEGMENT EQUAL TO AN ARBITRARY STRAIGHT-LINE SEGMENT Given an arbitrary point and an arbitrary segment, it is possible to construct a segment with one end- point being the previously given point such that its length is equal to that of the arbitrary segment. PROOF. Let A be an arbitrary point on the plane, and let BC be an arbi- trary segment. We wish to construct a segment with point A as an endpoint and with length equal to that of BC. On AB, construct the equilateral triangle AABD [1.1]. With B as the center and BC as the radius, construct the circle oECH [Postulate 1.3]. Extend DB to meet the circle oECH at E [Postulate 1.2]. With D as the center and DE as radius, construct the circle oEFG [Postulate 1.3]. Extend DA to meet oEFG at F. We claim that AF = BC. FIGURE 1.5.2. [1.2] partially constructed 1.5. PROPOSITIONS FROM BOOK I: 1-26 41 Because D is the center of the circle oEFG, DF = DE [Def. 1.32]. Because ADAB is an equilateral triangle, DA = DB [Def. 1.21]. Removing DA from DF and DB from DE, we have that AF = BE [Axiom 1.3]; that is, DF \u2014 DA = DE – DB 1 => AF = BE DE \u2014 DB = BE ) Again, because B is the center of the circle oECH, BC = BE. Since AF = BE, by [Axiom 1.1] we have that AF = BC. Therefore from the given point A, the segment AF has been constructed such that AF = BC. \u25a1 Exercises. 1. Prove [1.2] when A is a point on BC. Proposition 1.3. CUTTING A STRAIGHT-LINE SEGMENT AT A GIVEN SIZE. Given two arbitrary segments which are unequal in length, it is possible to cut the larger segment such that one of its two subsegments is equal in length to the smaller segment. PROOF. Let the arbitrary segments CG and AB be constructed such that AB > CG. We wish to show that AB may be subdivided into segments AE and EB where AE = CG. From the point A, construct the segment AD such that AD = CG [1.2]. With A as the center and AD as radius, construct the circle oEDF [Postulate 1.3] which cuts AB at E. We claim that AE = CG. 1.5. PROPOSITIONS FROM BOOK I: 1-26 \u2022\u00a3 . G 42 Figure 1.5.4. [1.3] Because A is the center of the circle oEDF, AE = AD [Def. 1.32]. Also, AD = CG by construction. By [Axiom 1.1], we have that AE = CG. \u25a1 COROLLARY. 1.4.1. Given an arbitrary segments and a ray, it is possible to cut the ray such that the resulting segment is equal in length to the arbitrary segment. Examination questions. 1. What previous problem is employed in the solution of this? 2. What postulate? 3. What axiom is employed in the demonstration? 4. Demonstrate how to extend the lesser of the two given segments until the whole extended segment is equal to the greater segment. PROPOSITION 1.4. THE “SIDE-ANGLE-SIDE” THEOREM FOR THE CON- GRUENCE OF TRIANGLES. If two pairs of sides of two triangles are equal in length and the corresponding interior angles are equal in measurement, then the triangles are congruent. PROOF. Suppose we have triangles AABC and ADEF such that: (a) the length of side AB of triangle AABC is equal in length to side DE of triangle ADEF, (b) the length of side AC of triangle AABC is equal in length to side DF of triangle ADEF, (c) the measure of the angle ABAC is equal in measure to the angle ZEDF. We wish to show that AABC ^ ADEF. 1.5. PROPOSITIONS FROM BOOK I: 1-26 43 C F = Figure 1.5.5. [1.4] Suppose that ABAC is applied to AEDF such that point A coincides with D, the segment AB coincides with segment DE, and point C stands on the same side of the segment AB as F is relative to DE. Because AB = DE, the point B coincides with point E. Because ZBAC = AEDF, AC coincides with DF; and since AC = DF by hypothesis, the point C coincides with point F. This proves that point B coincides with point E. Hence two points of the segment BC coincide with two points of the segment EF. And since two segments cannot enclose a space, BC must coincide with EF. Hence the triangles agree in every respect: BC = EF, ZABC = ZDEF, ZBCA = ZEFD, from which it follows that AABC ^ ADEF. \u25a1 Examination questions. 1. How many assumptions do we make in the hypothesis of this proposi- tion? (Ans. 3. Name them.) 2. How many in the conclusion? Name them. 3. What technical term is applied to figures which agree in everything but position? 4. What is meant by superposition? 5. What axiom is made use of in superposition? 6. How many parts in a triangle? (Ans. 6, three sides and three angles.) 7. When it is required to prove that two triangles are congruent, how many parts of one must be given equal to corresponding parts of the other? (Ans. In general, any three except the three angles. This will be established in [1.8] and [1.26], both of which use [1.4].) 8. What property of two segments having two common points is quoted in this proposition? (Ans. They must coincide.) Exercises. Prove the following: 1.5. PROPOSITIONS FROM BOOK I: 1-26 44 1. The line that bisects the vertical angle of an isosceles triangle bisects the base perpendicularly. 2. If two adjacent sides of a quadrilateral are equal and the diagonal bi- sects the angle between them, then their remaining sides are equal. 3. If two segments stand perpendicularly to each other and if each bisects the other, then any point in either segment is equally distant from the end- points of the other segment. 4. If equilateral triangles are constructed on the sides of any triangle, the distances between the vertices of the original triangle and the opposite vertices of the equilateral triangles are equal. (This should be proven after studying [1.32].) PROPOSITION 1.5. ISOSCELES TRIANGLES I. Suppose a given triangle is isosceles. Then 1) if the sides of the triangle other than the base are extended, the angles under the base are equal to each other, 2) the angles at the base are equal to each other. PROOF. Construct the triangle AABC such that sides AB = AC and de- note side BC as the triangle’s base. Extend the side AB to the segment BD and the side AC to the segment CE. We claim that the angles at the base {AABC, ZACB) are equal in measure to one another and that the external angles below the base (ZDBC, ZECB) are also equal in measure. Figure 1.5.6. [1.5] We prove each claim separately: 1. Let F be any point on the segment BD except for its endpoints. From AE, choose a point G such that CG = BF [1.3]. Join BG, CF [Postulate 1.1]. 1.5. PROPOSITIONS FROM BOOK I: 1-26 45 Because AF = AG by construction and AC = AB by hypothesis, sides AF, AC in triangle AFAC are equal in length to sides AG, AB in triangle AGAB. Also, the angle ZBAC is the interior angle to both pairs of sides in each triangle. By [1.4], AFAC ^ AGAB. Again, because AF = AG by construction and AB = AC by hypothesis, the remaining segments BF and CG are equal in length (or BF = CG) [Axiom 1.3]. Notice that ZAFC = ZBFC and ZAGB = ZCGB. Since we have shown that FB = CG, FC = GB, and ZAFC = ZAGB, by [1.4] we have that triangles AFBC ^ AGCB. Thus, ZDBC = ZFBC = ZGCB = ZECB which are the angles under the base. 2. We also have that ZFCB = ZGBC and ZFCA = ZGBA by the above. Since ZFCA = ZFCB + ZACB and ZGBA = ZGBC + ZABC, we obtain ZACB = ZABC which are the angles at the base. \u25a1 Observation: The great difficulty which beginners have with this proposi- tion is due to the fact that the two triangles AACF, AABG overlap each other. A teacher or tutor should graph these triangles separately and point out the corresponding parts: AF = AG, AC = AB, and ZFAC = ZGAB. By [1.4], we have that ZACF = ZABG, ZAFC = ZAGB. The student should also be shown how to apply one of the triangles to the other so as to bring them into coincidence. COROLLARY. 1. Every equilateral triangle is equiangular. Exercises. 1. Prove that the angles at the base are equal without extending the sides. Do the same by extending the sides through the vertex. 2. Prove that the line joining the point A to the intersection of the segments CF and BG is an Axis of Symmetry of AABC. 3. If two isosceles triangles stand on the same base, either on the same or on opposite sides of it, the line joining their vertices is an Axis of Symmetry of the figure formed by them. (Hint: this follows almost immediately from the proof of #2.) 4. Show how to prove this proposition by assuming as an axiom that every angle has a bisector. 1.5. PROPOSITIONS FROM BOOK I: 1-26 46 5. Prove that each diagonal of a rhombus is an Axis of Symmetry of the rhombus. (Hint: this follows almost immediately from the proof of #3.) 6. If three points are taken on the sides of an equilateral triangle (one on each side and at equal distances from the angles), then the segments joining them form a new equilateral triangle. PROPOSITION 1.6. ISOSCELES TRIANGLES II. If a given triangle has two equal angles, then the sides opposite the two angles are equal in length (i.e., the triangle is isosceles). Figure 1.5.7. [1.6] PROOF. Let A ABC be a triangle such that ZABC = ZACB. To show that side AB = AC, we will use a proof by contradiction^} Without loss of generalit}{^J suppose side AB > AC. On AB, construct a point D such that BD = CA [1.3] and join points C and D [Axiom 1.1]. Then in triangles ADBC, AACB, we have that BD = AC with BC being a common side to both. Therefore, the two sides DB, BC in triangle ADBC are equal to the two sides AC, CB in AACB. Also, we have that ZDBC = ZABC by hypothesis. By [1.4], we have that ADBC = AACB, a contradiction, since AB = AD + DB. It follows that AC, AB are not unequal; that is, AC = AB. \u25a1 Examination questions. 1. What is the hypothesis in this proposition? 2. What proposition is this the converse of? 9 That is, we will show that the statement “A triangle with two equal angles has unequal opposite sides” is false. See also https : \/\/en.wikipedia. org\/wiki\/Proof _by_contradiction 10 This term is used before an assumption in a proof which narrows the premise to some special case; it is implied that either the proof for that case can be easily applied to all others or that all other cases are equivalent. Thus, given a proof of the conclusion in the special case, it is trivial to adapt it to prove the conclusion in all other cases. http:\/\/en.wikipedia.org\/wiki\/Without_ |loss_of .generality! 1.5. PROPOSITIONS FROM BOOK I: 1-26 47 3. What is the inverse of this proposition? 4. What is the inverse of [1.5]? 5. What is meant by an indirect proof? (Ans. A proof by contradiction.) 6. How does Euclid generally prove converse propositions? 7. What false assumption is made in order to prove the proposition? 8. What does this false assumption lead to? COROLLARY. 1. A triangle is isosceles if and only if the angles at its base are equal. PROPOSITION 1.7. UNIQUENESS OF TRIANGLES. Suppose that we have two distinct triangles which share a common base. Also suppose that at one endpoint of the base we have that the two sides which connect to this vertex are equal in length. It follows that the lengths of the sides of the triangles which are connected to the other endpoint of the base are unequal in length. PROOF. Construct distinct triangles AADB, AACB which share the base AB. Suppose that AC = AD where each side shares the common endpoint A. We claim that BC ^ BD. Figure 1.5.8. [1.7], case 1 We prove this claim in two cases. Case 1: Let the vertex of each triangle lie outside the interior of the other triangle (i.e., such that D does not lie inside AACB and C does not lie inside AADB). Join CD. Because AD = AC by hypothesis, AACD is isosceles. By [1.5], ZACD = ZADC. However, ZADC > ZBDC since ZADC = ZADB + ZBDC [Axiom 1.9]. Therefore ZACD > ZBDC. Since ZBCD = ZBCA + ZACD, we also have that ZBCD > ZBDC. Now if BD = BC, we would have that ZBCD = ZBDC [1.5]; however, we have shown that ZBCD ^ ZBDC. Hence, BD ^ BC. 1.5. PROPOSITIONS FROM BOOK I: 1-26 48 Figure 1.5.9. [1.7], case 2 Case 2: Wlogf^J let the vertex of the triangle AADB fall within the inte- rior of AACB. Extend side AC to segment CE and side AD to segment DF. Because AC = AD by hypothesis, the triangle AACD is isosceles, and by [1.5] ZECD = ZFDC (i.e., the external angles at the other side of the base CD are equal). Notice that ZECD > ZBCD since ZECD = ZECB + ZBCD [Axiom 1.9]. Therefore ZFDC > ZBCD. Since ZBDC = ZBDF + ZFDC, we have that ZBDC > ZBCD. Now if BC = BD, we would have that ZBDC = ZBCD [1.5]; however, we have shown that ZBCD ^ ZBDC. Hence BD ^ BC. \u25a1 COROLLARY. 1.7.1. Triangles which have no sides of equal length are dis- tinct. Examination questions. 1. What use is made of [1.7]? (Ans: As a lemma to [1.8].) 2. In the demonstration of [1.7], the contrapositive of [1.5] occurs. Show where. PROPOSITION 1.8. THE “SIDE-SIDE-SIDE” THEOREM FOR THE CON- GRUENCE OF TRIANGLES. If three pairs of sides of two triangles are equal in length, then the triangles are congruent. PROOF. If two distinct triangles (A ABC, ADEF) have two sides (AB, AC) that are respectively equal to two sides of the other (DE, DF) where the base of one triangle (BC) equals the base of the other (EF), we claim that the two triangles are congruent. “^An abbreviation for “without loss of generality”. |http : \/\/en . wikipedia . org\/wiki\/Withoutr |loss_of .generality! 1.5. PROPOSITIONS FROM BOOK I: 1-26 49 F Figure 1.5.10. [1.8] Let the triangle AABC be applied to ADEF such that point B coincides with E and the side BC coincides with the side EF. Because BC = EF, the point C coincides with point F. If the vertex A falls on the same side of EF as vertex D, then the point A must coincide with D. Otherwise, it must take a different position, such as point G. We then have EG = BA and BA = ED (by hypothesis). By [Axiom 1.1], EG = ED. Similarly, FG = FD, a contradiction since the triangles are distinct [1.7]. Hence the point A must coincide with D, and so the three angles of one triangle are respectively equal to the three angles of the other (specifically, ZABC = ZDEF, ZBAC = ZEDF, and ABC A = ZEFD). Therefore, AABC ^ ADEF. \u25a1 This proposition is the converse of [1.4] and is the second case of the con- gruence of triangles in the Elements. Philo’s Proof: PROOF. Let the equal bases be applied as in the foregoing proof, but let the vertices fall on the opposite sides of the base. Let ABGC be a “copy” of AEDF. Join AG. Because BG = BA, we have that ZBAG = ZBGA. Similarly, ZCAG = ZCGA. FIGURE 1.5.11. Philo’s Proof of [1.8] 1.5. PROPOSITIONS FROM BOOK I: 1-26 50 Notice that ZBGA + ZCGA = ZBGC ii ZBAG + ZCAG = ZBAC Hence, ZBAC = ZBGC. Since ABGC = AEDF by construction, we have that ABAC = AEDF. \u25a1 Proposition 1.9. BISECTING A RECTILINEAR ANGLE. It is possible to bisect an angle. Figure 1.5.12. [1.9] PROOF. Take any point D on the ray AB. Take the point E on AC such that AE = AD [1.3]. Join DE [Postulate 1.1] and, on the opposite side of point A, construct the equilateral triangle ADEF [1.1]. Join AF. We claim that AF bisects ZBAC. Notice that triangles ADAF [AEAF share the side AF. Given that AD = AE by construction, we have that the two sides DA, AF are respectively equal to the two sides EA, AF. Also, DF = EF because they are the sides of an equilateral triangle [Def. 1.21]. By [1.8], ZDAF = ZEAF. Since ZBAC = ZDAF + ZEAF, ZBAC is bisected by AF. \u25a1 COROLLARY. 1. The line AF is an Axis of Symmetry of the figure ABCF, AAED, and segment DE. 1.5. PROPOSITIONS FROM BOOK I: 1-26 51 COROLLARY. 2. In [1.9], AB and AC may be constructed as either lines, rays, or segments with point A as the vertex, mutatis mutandis. Examination questions. 1. Why does Euclid construct the equilateral triangle on the side opposite of A? 2. If the equilateral triangle were constructed on the other side of DE, in what case would the construction fail? Exercises. 1. Prove [1.9] without using [1.8]. (Hint: the proof follows almost immedi- ately from [1.5, #2].) 2. Prove that AF _L DE. (Hint: the proof follows almost immediately from [1.5, #2].) 3. Prove that any point on AF is equally distant from the points D and E. Proposition 1.10. BISECTING A STRAIGHT-LINE SEGMENT. It is possible to bisect an arbitrary segment; that is, it is possible to locate the mid- point of a segment. PROOF. We wish to bisect the segment AB. :. Figure 1.5.13. [1.10] Construct an equilateral triangle AACB on segment AB [1.1]. Bisect ZACB by the segment CD [1.9], intersecting AB at point D. We claim that AB is bisected at D. The two triangles AACD, ABCD have sides AC, BC such that AC = BC (since each are sides of an equilateral triangle) and also share side CD in common. Therefore, the two sides AC, CD in one triangle are equal to the two sides BC, CD in the other. We also have that ZACD = ZBCD by construction. By [1.4], we have that AD = DB. Since AB = AD + DB, it follows that AB is bisected at D. \u25a1 1.5. PROPOSITIONS FROM BOOK I: 1-26 52 Exercises. 1. Bisect a segment by constructing two circles. 2. Extend CD to a line. Prove that every point equally distant from the points A, B are points in the line CD. Proposition 1.11. CONSTRUCTING A PERPENDICULAR STRAIGHT- LINE SEGMENT TO A LINE I. It is possible to construct a segment at a right angle to a given line from an arbitrary point on the line. PROOF. Construct the line AB containing the point C. We wish to con- struct the segment CF such that AB _L CF. On AC, take any point D and choose some point E on AC such that CE \u2014 CD [1.3]. Construct the equilateral triangle ADFE on DE [1.1] and join CF. We claim that AB _L CF. The two triangles ADCF, AECF have CD = CE by construction and CF in common; therefore, the two sides CD, CF in one triangle are respectively equal to the two sides CE, CF in the other, and DF = EF since they are the sides of an equilateral triangle [Def. 1.21]. By [1.8], ZDCF = ZECF. Since these are adjacent angles, by [Def. 1.13] each is a right angle, and so AB _L CF at point C. \u25a1 COROLLARY. 1. [1.11] holds whenAB is a segment or ray and \/or when CF is a straight line or a ray, mutatis mutandis. Exercises. 1. Prove that the diagonals of a lozenge bisect each other perpendicularly. 2. Prove [1.11] without using [1.8]. (Hint: The proof follows from the result 3. Find a point on a given line that is equally distant from two given points. Figure 1.5.14. [1.11] of [1.9, #2].) 1.5. PROPOSITIONS FROM BOOK I: 1-26 53 4. Find a point on a given line such that if it is joined to two given points on opposite sides of the line, then the angle formed by the connecting segment is bisected by the given line. (Hint: similar to the proof of #3.) 5. Find a point that is equidistant from three given points. (Hint: you are looking for the circumcenter of the triangle formed by the pointful ) Proposition 1.12. CONSTRUCTING A PERPENDICULAR STRAIGHT- LINE SEGMENT TO A LINE II. Given an arbitrary straight line and an arbi- trary point not on the line, we may construct a perpendicular segment from the point to the line. PROOF. We wish to construct a perpendicular segment to a given line AB from a given point C which is not on AB. Figure 1.5.15. [1.12] Take any point D which is not on AB and stands on the opposite side of AB from C. Construct the circle oFDG with C as its center and CD as radius [Postulate 1.3] which intersects AB at the points F and G. Bisect the segment FG at H [1.10] and join CH [Postulate 1.1]. We claim that CH _L AB. To see this, join CF, CG. Then the two triangles AFHC, AGHC have sides FH = GH by construction and side HC in common. We also have that CF = CG since both are radii of oFDG [Def. 1.32]. Therefore, ZCHF = ZCHG [1.8]. Since these are adjacent angles, by [Def. 1.13] each is a right angle, and so CH _L AB. \u25a1 COROLLARY. 1. [1.12] holds when CH and lor AB are replaced by rays, mutatis mutandis. http : \/\/www .mathopenref . com\/trianglecircumcenter .html 1.5. PROPOSITIONS FROM BOOK I: 1-26 54 Exercises. 1. Prove that circle oFDG cannot meet AB at more than two points. 2. If one angle of a triangle is equal to the sum of the other two, prove that the triangle can be divided into the sum of two isosceles triangles and that the length of the base is equal to twice the segment from its midpoint to the opposite angle. Proposition 1.13. ANGLES AT INTERSECTIONS OF STRAIGHT LINES. If a line intersects another line at one and only one point, the lines will either make two right angles or two angles whose sum equals two right angles. PROOF. If the line AB intersects the line CD at one and only one point (B), we claim that either ZABC and ZABD are right angles or the sum ZABC + ZABD equals two right angles. Figure 1.5.16. [1.13] (a), (f3) If AB _L CD as in Fig. 1.5.16(a), then A ABC and ZABD are right angles. Otherwise, construct BE _L CD [1.11]. Notice that ZCBA = AC BE + ZEBA [Def. 1.11]. Adding the measure of ZABD to each side of the equality, we obtain that ZCBA + ZABD = ZCBE + ZEBA + ZABD Similarly, we have that ZCBE + ZEBD = ZCBE + ZEBA + ZABD Since quantities which are equal to the same quantity are equal to one another, we have that ZCBA + ZABD = ZCBE + ZEBD Since ZCBE, ZEBD are right angles, we have that ZCBA + ZABD equals the sum of two right angles. \u25a1 1.5. PROPOSITIONS FROM BOOK I: 1-26 55 An alternate proof: PROOF. Denote ZEBA by 0. We then have that ZCBA = right angle + 0 ZABD = right angle -9 => ZCBA + ZABD = right angle \u25a1 COROLLARY. 1. The above proposition holds when the straight lines are replaced by segments and \/or rays, mutatis mutandis. COROLLARY. 2. The sum of two supplemental angles equals two right an- gles. COROLLARY. 3. Two distinct straight lines cannot share a common seg- ment. COROLLARY. 4. The bisector of any angle bisects the corresponding re- entrant angle. COROLLARY. 5. The bisectors of two supplemental angles are at right an- gles to each other. COROLLARY. 6. The angle ZEBA is half the difference of the angles ZCBA, ZABD. PROPOSITION 1.14. RAYS TO STRAIGHT LINES. If at the endpoint of a ray there exists two other rays standing on opposite sides of that ray such that the sum of their adjacent angles is equal to two right angles, then these two rays form one line. PROOF. Construct the ray BA with endpoint B. Suppose at B there exists two other rays BC and BD which stand on opposite sides of BA such that the sum of their adjacent angles, ZCBA + ZABD, equals two right angles. We claim that BC \u00a9 BD = CD where CD is a line. 1.5. PROPOSITIONS FROM BOOK I: 1-26 56 Figure 1.5.17. [1.14] Suppose instead that the rays BC and BE form the straight line CE. Since CE is a line and BA stands on it, the sum ZCBA + ZABE equals two right angles [1.13]. Also by hypothesis, the sum ZCBA + ZABD equals two right angles. Therefore, ZCBA + ZABE = ZCBA + ZABD Removing the angle in common, ZCBA, we have that ZABE = ZABD. However, ZABD = ZABE + ZEBD, a contradiction [Axiom 1.9]. Hence, BC \u00a9 \u00a3L> = CD where CD is a line. \u25a1 COROLLARY, i. If at a point on a straight line, segment, or ray, two seg- ments on opposite sides of the line make the sum of the adjacent angles equal to two right angles, these two segments also form a single segment. PROPOSITION 1.15. OPPOSITE ANGLES ARE EQUAL. If two lines inter- sect one another at one point, their opposite angles are equal. PROOF. If two straight lines AB, CD intersect one another at one point, E, then their opposite angles are equal (ZCEA = ZDEB and ZBEC = ZAED). Figure 1.5.18. [1.15] Because the line AB intersects CD at E, the sum ZCEA + ZAED equals two right angles [1.13]. Because the line CD intersects AB at the point E, the sum ZCEA + ZBEC also equals two right angles. Therefore, ZCEA + ZAED = ZCEA + ZBEC 1.5. PROPOSITIONS FROM BOOK I: 1-26 57 Removing the common angle ZCEA, we have that ZAED = ZBEC. By an analogous method, we also obtain that ZCEA = ZDEB. \u25a1 An alternate proof: PROOF. Because opposite angles share a common supplement, they are equal. \u25a1 COROLLARY. 1. [1.15] holds when either one or both of the two straight lines are replaced either by segments or by rays, mutatis mutandis. Examination questions for [1.13]-[1.15]. 1. What problem is required in Euclid’s proof of [1.13]? 2. What theorem? (Ans. No theorem, only the axioms.) 3. If two lines intersect, how many pairs of supplemental angles do they make? 4. What is the relationship between [1.13] and [1.14]? 5. What three lines in [1.14] are concurrent? 6. What caution must be taken as we prove [1.14]? 7. State the converse of Proposition [1.15] and prove it. 8. What is the subject of [1.13], [1.14], [1.15]? (Ans. Angles at a point.) Proposition 1.16. THE EXTERIOR ANGLE IS GREATER THAN EI- THER OF THE NON-ADJACENT INTERIOR ANGLES. If any side of a tri- angle is extended, the exterior angle is greater than either of the non-adjacent interior angles. PROOF. Construct AABC. Wlog, we extend side BC to the segment CD. We claim that the exterior angle ZACD is greater than either of the interior non-adjacent angles ZABC, ZBAC. 1.5. PROPOSITIONS FROM BOOK I: 1-26 58 r S \/ \/ \/ \/ Figure 1.5.19. [1.16] Bisect AC at point E [1.10] and join BE [Postulate 1.1]. Extend BE to EF such that BE = EE [1.3]. Join CF. Because EC = EA by construction, the triangles ACEF, AAEB have the sides CE, EF in one equal to the sides AE, EB in the other. We also have that ZCEF = ZAEB [1.15]. By [1.4], the angle ZECF = ZEAB. However, ZACD is greater than ZECF since ZACD = ZECF + ZFCD. Therefore, the angle ZACD > ZEAB = ZBAC. Similarly, it can be shown that if side AC is extended to segment CG, then the exterior angle ZBCG > ZABC. But ZBCG = ZACD [1.15]. Hence, ZACD is greater than either of the interior non-adjacent angles ZABC or ZBAC. \u25a1 COROLLARY. 1. The sum of the three interior angles of the triangle ABCF is equal to the sum of the three interior angles of the triangle A ABC. COROLLARY. 2. The area of ABCF is equal to the area of A ABC, which we will write as ABCF = AABC. COROLLARY. 3. The lines BA and CF, if extended, cannot meet at any finite distance. For, if they met at any finite point X, the triangle AC AX would have an exterior angle ZBAC equal to the interior angle ZACX. Exercise. 1. Prove [1.16, Cor. 3] using a proof by contradiction. Proposition 1.17. THE SUM OF TWO ANGLES OF A TRIANGLE. The sum of two angles of a triangle is less than that of two right angles. 1.5. PROPOSITIONS FROM BOOK I: 1-26 59 PROOF. We claim that the sum of any two angles of a triangle AABC is less than the sum of two right angles (wlog, we choose ZABC and ZBAC). Figure 1.5.20. [1.17] Extend side BC to the segment CD. By [1.16], the exterior angle ZACD is greater than ZABC. To each, add the angle ZACB, and we obtain ZACD + ZACB > ZABC + ZACB. However, ZACD + ZACB equals two right angles [1.13]. Therefore, ZABC + ZACB is less than two right angles. Similarly, we may show that the sums ZABC + ZBAC and ZACB + Z# AC are each less than two right angles. A similar argument follows on AEFG, mutatis mutandis. \u25a1 COROLLARY. 1. Every triangle has at least two acute angles. COROLLARY. 2. If two angles of a triangle are unequal, the lesser is acute. Exercise. 1. Prove [1.17] without extending a side. (Attempt after completing Chap- ter 1. Hint: use parallel line theorems.) PROPOSITION 1.18. ANGLES AND SIDES IN A TRIANGLE I. In a trian- gle, if one side is longer than another, then the angle opposite to the longer side is greater in measure than the angle opposite to the shorter side. PROOF. Suppose we have AABC with sides AB, AC such that AC > AB. We claim that the angle opposite AC (ZABC) is greater in measure than the angle opposite AB (ZACB). 1.5. PROPOSITIONS FROM BOOK I: 1-26 60 Figure 1.5.21. [1.18] From AC, cut off AD such that AD = AB [1.3]. Join BD [Postulate 1.1]. It follows that AABD is isosceles; therefore, ZADB = ZABD. Now ZADB > ZACB [1.16], and so ZABD > ZACB. Since ZABC = ZABD + ZCBD, we also have that ZABC > ZABD from which it follows that ZABC > ZACB. \u25a1 An alternate proof: PROOF. With A as the center and with the shorter side AB as radius, con- struct the circle oBED which intersects BC at point E. FIGURE 1.5.22. [1.18], alternate proof Join AE. Since AB = AE, ZAEB = ZABE; however, ZAEB > ZACB [1.16]. Therefore, ZABE > ZACB. \u25a1 Exercises. 1. Prove that if two of the opposite sides of a quadrilateral are respectively the greatest and the least sides of the quadrilateral, then the angles adjacent to the least are greater than their opposite angles. 2. In any triangle, prove that the perpendicular from the vertex opposite the side which is not less than either of the remaining sides falls within the triangle. 1.5. PROPOSITIONS FROM BOOK I: 1-26 61 PROPOSITION 1.19. ANGLES AND SIDES IN A TRIANGLE II. In a tri- angle, if one angle is greater in measure than another, then the side opposite the greater angle is longer than the side opposite the shorter angle. PROOF. Construct AABC with sides AB, AC. We claim that if ZABC > ZACB, then AC > AB. Figure 1.5.23. [1.19] If AC is not longer than AB, then either AC = AB or AC < AB. 1. If AC = AB, AACB is isosceles and ZACB = ZABC [1.5]. This contra- dicts our hypothesis, and so AC ^ AB. 2. If AC < AB, we have that ZACB AB. \u25a1 COROLLARY. 1. In a triangle, greater (lesser) sides stand opposite the greater (lesser) angles and greater (lesser) angles stand opposite the greater (lesser) sides. Exercises. 1. Prove this proposition by a direct demonstration. 2. Prove that a segment from the vertex of an isosceles triangle to any point on the base is less than either of the equal sides but greater if the base is extended and the point of intersection falls outside of the triangle. 3. Prove that three equal segments cannot be constructed from the same point to the same line. 4. If in [1.16], Fig 1.5.19, AB is the longest side of the A ABC, then BF is the longest side of AFBC and ZBFC is less than half of ZABC. 5. If AABC is a triangle such that side AB AC, then a segment AG, constructed from A to any point G on side BC, is less than AC. 1.5. PROPOSITIONS FROM BOOK I: 1-26 62 Proposition 1.20. THE SUM OF THE LENGTHS OF ANY PAIR OF SIDES OF A TRIANGLE. In a triangle, the sum of the lengths of any pair of sides is greater than the length of the remaining side. PROOF. We claim that the sum of any two sides (BA, AC) of a triangle A ABC is greater than the third (BC). Figure 1.5.24. [3.20] Extend BA to the segment AD [Postulate 1.2] such that AD = AC [1.3]. Join CD, constructing AACD. Because AD = AC, ZACD = ZADC [1.5]. Since ZBCD = ABC A + AACD, ZBCD > ZADC = ZBDC. It follows that the side BD > BC [1.19]. Noticing that AD = AC and BA + AD = BA + AC BD = BA + AC we obtain that BA + AC > BC. \u25a1 Alternatively: PROOF. Bisect the angle ZBAC by AE [1.9]. Then the angle ABE A is greater than ZEAC. However, ZEAC = ZEAB by construction. Therefore, the angle ABE A > ZEAB. It follows that BA > BE [1.19]. Similarly, AC > EC. It follows that BA + AC > BE + EC = BC. \u25a1 Exercises. 1. In any triangle, the difference between the lengths of any two sides is less than the length of the third. 2. Any side of any polygon is less than the sum of the remaining sides. 1.5. PROPOSITIONS FROM BOOK I: 1-26 63 3. The perimeter of any triangle is greater than that of any inscribed tri- angle and less than that of any circumscribed triangle. (See also [Def. 4.1].) 4. The perimeter of any polygon is greater than that of any inscribed (and less than that of any circumscribed) polygon of the same number of sides. 5. The perimeter of a quadrilateral is greater than the sum of its diagonals. 6. The sum of the lengths of the three medians of a triangle is less than 3\/2 times its perimeter. PROPOSITION 1.21. TRIANGLES WITHIN TRIANGLES. In an arbitrary triangle, if two segments are constructed from the vertexes of its base to a point within the triangle, then 1) the sum of these inner sides will be less than the sum of the outer corre- sponding sides (i.e., the outer sides excluding the base); 2) these inner sides will contain a greater angle than the corresponding sides of the outer triangle. PROOF. If two segments (BD, CD) are constructed to a point (D) within a triangle ( AABC) from the endpoints of its base (BC), we claim that: 1. BA + AC > BD + DC 2. ZBDC > ABAC Construct AABC and ABDC as in Fig. 1.5.25(a). Figure 1.5.25. [3.21] (a), (f3) 1. Extend BD to meet AC at point E [Postulate 1.2]. In triangle ABAE, we have that BA + AE > BE [1.20], from which it follows that BA + AC = BA + AE + EC > BE + EC Similarly, in ADEC, we have that DE + EC > DC, from which it follows that BE + EC = BD + DE + EC> BD + DC From these two inequalities, we obtain that BA + AC > BD + DC. 1.5. PROPOSITIONS FROM BOOK I: 1-26 64 2. Consider ADEC. By [1.16], we have that ZBDC > ZBEC. Similarly in AABE, ZBEC > ZBAE. It follows that ZBDC > ZBAE = ZBAC. \u25a1 An alternative proof to part 2 that does not extend sides BD, DC: PROOF. Construct AABC and ABDC as in Fig. 1.5.25(\/3), joining AD and extending it to meet BC at point F. Consider ABDA and ACDA. By [1.16], ZBDF > ZBAF and ZFDC > ZFAC. Since ZBDF + ZFDC = ZBDC ZBAE + ZFAC = ZBAC we have that ZBDC > ZBAC. \u25a1 Exercises. 1. The sum of the side lengths constructed from any point within a triangle to its angular points is less than the length of the triangle’s perimeter. Figure 1.5.26. [1.21, #2] 2. If a convex polygonal line ABCD lies within a convex polygonal line AMND terminating at the same endpoints, prove that the length of the former is less than that of the latter. Proposition 1.22. CONSTRUCTION OF TRIANGLES FROM ARBITRARY SEGMENTS. It is possible to construct a triangle whose three sides are respec- tively equal to three arbitrary segments whenever the sum of every two pairs of segments is greater than the length of the remaining segment. PROOF. Let AR, BS, and CT be arbitrary segments which satisfy our hy- pothesis. 1.5. PROPOSITIONS FROM BOOK I: 1-26 65 Figure 1.5.27. [1.22] Take the ray DE and cut off the segments DF = AR, FG = BS, and GH = CT [1.3]. With F as the center and FD as radius, construct the circle oKDL [Postulate 1.3]. With G as the center and GH as radius, construct the circle oKHL where K is a point of intersection between the circles oKDL and oKHL. Join KF, KG. We claim that AKFG is the required triangle. Since F is the center of oKDL, FK = FD. Since FD = AR by construction, FK = AR [Axiom 1.1]. Also by construction, we have that GK = CT and FG = BS. Hence, the three sides of the triangle AKFG are respectively equal to the three segments AR, BS, and CT. \u25a1 Examination questions. 1. What is the reason for our condition that the sum of every two of the given segments must be greater than the length of the third? 2. Under what conditions would the circles fail to intersect? Exercises. 1. Prove that when the above condition is fulfilled that the two circles must intersect. 2. If the sum of two of the segments equals the length of the third, would the circles meet? Prove that they would not intersect. Proposition 1.23. CONSTRUCTING AN ANGLE EQUAL TO AN AR- BITRARY RECTILINEAR ANGLE. It is possible to construct an angle equal to an arbitrary angle on an endpoint of a segment. 1.5. PROPOSITIONS FROM BOOK I: 1-26 66 PROOF. Construct an arbitrary angle ZDEF from the rays ED, EF. From a given point A on a given segment AB, we wish to construct an angle equal to ZDEF. Figure 1.5.28. [1.23] Join DF and construct the triangle ABAC whose sides are respectively equal to those of DEF; specifically let AB = ED, AC = EF, and CB = FD [1.22]. By [1.8], ABAC = ZDEF. \u25a1 Exercises. 1. Construct a triangle given two sides and the angle between them. 2. Construct a triangle given two angles and the side between them. 3. Construct a triangle given two sides and the angle opposite one of them. 4. Construct a triangle given the base, one of the angles at the base, and the sum or difference of the sides. 5. Given two points, one of which is in a given line, find another point on the given line such that the sum or difference of its distances from the former points may be given. Show that two such points may be found in each case. Proposition 1.24. ANGLES AND SIDES IN A TRIANGLE III. If in two triangles we have two pairs of sides in each triangle respectively equal to the other where the interior angle in one triangle is greater in measure than the interior angle of the other triangle, then the remaining sides of the triangles will be unequal in length; specifically, the triangle with the greater interior angle will have a greater side than the triangle with the lesser interior angle. PROOF. Construct two triangles AABC, ADEF where two sides of one (AB, AC) are respectively equal to two sides of the other (DE, DF) but the in- terior angle of AABC (ZBAC) is greater than the interior angle of the ADEF UEDF). We claim that the base of AABC (BC) is longer than the base of ADEF (EF). (Or, if AB = DE, AC = DF, and ZBAC > ZEDF, then BC > EF.) 1.5. PROPOSITIONS FROM BOOK I: 1-26 67 Figure 1.5.29. [1.24] Construct point G on AC such that ZBAG = ZEDF. Wlog, suppose that AC > AB. Because AB ? AC, AG ZBCH since ZACH = ZBCH+ZBCA. It follows that ZAHC > ZBCH. And since ZBHC = ZBHA+ ZAHC, we also have that ZBHC > ZBCH. By [1.19], the greater angle stands opposite to the longer side, and so BC > BH. Since BH = EF, it follows that BC > EF. \u25a1 Alternatively, the concluding part of this proposition may be proved with- out joining CH. PROOF. Construct the triangles as above. We have that BG + GH > BH [1.20] and AG + GC > AC [1.20] BC + AH > BH + AC Since AH = AC by construction, we have that BC > BH = EF. \u25a1 Another alternative: PROOF. In A ABC, bisect the angle ZCAH by AO and join OH. Now in AC AO, AH AO we have the sides CA, AO in one triangle equal to the sides AH, AO in the other where the interior angles are equal. By [1.4], OC = OH. 1.5. PROPOSITIONS FROM BOOK I: 1-26 68 It follows that BO + OH = BO + OC = \u00a3C. But BO + OH > BH [1.20]. Therefore, BO BH = EF. \u25a1 Exercises. 1. Prove this proposition by constructing the angle ZABH to the left of AB. 2. Prove that the angle ABC A > ZEFD. Proposition 1.25. ANGLES AND SIDES IN A TRIANGLE IV. If in two triangles we have two pairs of sides in each triangle respectively equal to the other where the remaining sides of the triangles are unequal, the interior angle of one triangle will be greater in measure than that of the other triangle; specif- ically, the triangle with the longer side will have a greater interior angle than the triangle with the shorter side. PROOF. If two triangles (AABC, ADEF) have two sides of one triangle (AB, AC) respectively equal to two sides of the other (DE, DF) where the base of one (BC) is greater than the base of the other (EF), the angle (ZBAC) contained by the sides of the triangle with the longer base ( A ABC) is greater in measure than the angle (ZEDF) contained by the sides of the other (ADEF). (Or, if AB = DE, AC = DF, and BC > EF, then ZBAC > ZEDF.) Figure 1.5.30. [1.25] If ZBAC ? ZEDF, then either ZBAC = ZEDF or ZBAC EF. Hence, ZBAC + ZEDF. 2. It ZEDF > ZBAC, then because the triangles ADEF, AABC have the two sides DE, DF of one respectively equal to the two sides AB, AC of the 1.5. PROPOSITIONS FROM BOOK I: 1-26 69 other, by [1.24], EF > BC, which contradicts our hypothesis that BC > EF. Therefore AEDF ^ ABAC. Since it is not the case that either ABAC = AEDF or ABAC AEDF. \u25a1 An alternate proof: PROOF. Construct AABC, ADEF as in the previous proof as well as the triangle AACG where sides AG = DE, GC = EF, CA = FD of the triangle ADEF [1.22]. Join BG. Because BC > EF by hypothesis, it follows that BC > GC. By [1.18], ABGC > AGBC. Construct ABGH = ZGBH [1.23] and join AH. By [1.6], BH = GH. FIGURE 1.5.31. [1.25], alternate proof Therefore in AABH, AAGH we have that AB = AG, BH = GH, and the side AH in common. By [1.8], ZBAH = AG AH. Since ABAC = ZBAH + AC AH, we also have that ABAC = AG AH + AC AH = AC AG + 2 \u2022 AC AH and therefore ABAC > AC AG. By [1.8], we have that AC AG = AEDF, and so ABAC > AEDF. \u25a1 COROLLARY. 1. In two triangles with two pairs of sides respectively equal to the other, the final sides are unequal if and only if the interior angles of the two pairs of sides are also unequal. Specifically, the statement that <( the triangle with the longer side contains a greater interior angle than the triangle with the shorter side" is equivalent to the statement that ( ZEFD, which contradicts our hypothesis that ZAEF = ZEFD. Hence, AB || CD. \u25a1 PROPOSITION 1.28. 1) PARALLEL LINES II. If a straight line intersecting two straight lines at one and only point point each makes the exterior angle equal to its corresponding interior angle, the two straight lines are parallel. 2) PARALLEL LINES III. If a straight line intersecting two straight lines at one and only point point makes two interior angles on the same side equal to two right angles, the two straight lines are parallel. PROOF. If a straight line (EF) intersects two straight lines (AB, CD) at one and only point point each such that the exterior angle (ZEGB) equals its corresponding interior angle (ZGHD), or if it makes two interior angles on the same side (ZBGH, ZDHG) equal to two right angles, then the two lines are parallel (AB \\ CD). :. : 1.6.4. [1.28] * We prove each claim separately: 1.6. PROPOSITIONS FROM BOOK I: 27-48 75 1. Suppose that ZEGB = ZGHD. Since the lines AB, EF intersect at G, ZAGH = ZEGB [1.15]. It follows that ZAGH = ZGHD. Since these are alternate angles, by [1.27], AB || CD. 2. Now suppose that the sum ZBGH + ZDHG equals two right angles. Since ZAGH and ZBGH are adjacent angles, the sum ZAGH + ZBGH equals two right angles [1.13]. If we remove the common angle ZBGH, we have that ZAGH = ZDHG, and these are alternate angles. By [1.27], AB || CD. \u25a1 PROPOSITION 1.29. PARALLEL LINES IV. If a straight line intersects two parallel straight lines at one and only one point each, then: 1) corresponding alternate angles are equal to each other, 2) exterior angles are equal to corresponding interior angles, 3) the sum of interior angles on the same side is equal to two right angles. PROOF. If a straight line EF intersects two parallel straight lines AB, CD at one and only one point each, we claim that: 1. alternate angles ZAGH, ZGHD are equal; 2. the exterior angle ZEGB equals its corresponding interior angles ZGHD; 3. the sum of the two interior angles ZHGB + ZGHD equals two right angles. Figure 1.6.5. [1.29] We prove each claim separately: 1. If ZAGH ^ ZGHD, one must be greater than the other. Wlog, suppose that ZAGH > ZGHD. Then we obtain the inequality ZAGH + ZBGH > ZGHD + ZBGH where ZAGH + ZBGH is equal to the sum of two right angles [1.13]. It follows that the sum ZGHD + ZBGH is less than two right angles. By [Axiom 1.12], the lines AB, CD meet at some finite distance, a contradiction – since they 1.6. PROPOSITIONS FROM BOOK I: 27-48 76 are parallel by hypothesis, they cannot meet at any finite distance. Hence, the angle ZAGH is not unequal to ZGHD; or, ZAGH = ZGHD. 2. Since ZEGB = ZAGH [1.15] and ZGHD = ZAGH by part 1 of this proof, it follows that ZEGB = ZGHD [Axiom 1.1]. 3. Since ZAGH = ZGHD by part 1 of this proof, we obtain ZAGH + ZHGB = ZGHD + ZHGB where the sum ZAGH + ZBGH equals the sum of two right angles. It follows that the sum ZGHD + ZHGB equals the sum of two right angles. \u25a1 Corollary. 1. EQUIVALENT STATEMENTS REGARDING PARAL- LEL LINES. Suppose two straight lines are intersected by a third straight line at one and only one point. The two straight lines are parallel if and only if any one of these three properties hold: 1) corresponding alternate angles are equal; 2) exterior angles equal their corresponding interior angles; 3) the sum of the interior angles on the same side are equal to two right angles. COROLLARY. 2. We may replace the straight lines in [1.29, Cor. 1] with segments or rays, mutatis mutandis. Exercises. Note: We may use [1.31] (that we may construct a straight line parallel to any given straight line) in the proofs of these exercises since the proof of [1.31] does not employ [1.29]. 1. Demonstrate both parts of [1.28] without using [1.27]. 2. If ZACD, ZBCD are adjacent angles, any parallel to AB will meet the bisectors of these angles at points equally distant from where it meets CD. 4. If any other secant is constructed through the midpoint O of any straight line terminated by two parallel straight lines, the intercept on this line made by the parallels is bisected at O. 5. Two straight lines passing through a point equidistant from two paral- lels intercept equal segments on the parallels. 6. The perimeter of the parallelogram, formed by constructing parallels to two sides of an equilateral triangle from any point in the third side, is equal to twice the side. 1.6. PROPOSITIONS FROM BOOK I: 27-48 77 7. If the opposite sides of a hexagon are equal and parallel, its diagonals are concurrent. 8. If two intersecting segments are respectively parallel to two others, the angle between the former is equal to the angle between the latter. For, if AB, AC are respectively parallel to DE, DF and if AC, DE intersect at G, the angles at points A, D are each equal to the angle at G [1.29]. PROPOSITION 1.30. TRANSITIVITY OF PARALLEL LINES. Straight lines parallel to the same straight line are also parallel to one another. PROOF. Construct straight lines AB, CD, EF such that AB || EF and CD || EF. We claim that AB || CD. Figure 1.6.6. [1.30] Construct any secant GHK. Since AB \\ EF, the angle ZAGH = ZGHF [1.29]. Similarly, the angle ZGHF = ZHKD [1.29]. By [Axiom 1.1], ZAGK = ZGKD, and by [1.27], we have that AB \\ CD. \u25a1 COROLLARY. 1. AB, CD, EF, and GK in [1.30] may be replaced by rays and lor straight-line segments, mutandis mutandis. Proposition 1.31. CONSTRUCTION OF A PARALLEL LINE. We wish to construct a straight line which is parallel to a given straight line and passes through a given point. PROOF. We wish to construct a straight line (CE) which is parallel to a given straight line (AB) and passes through a given point (C). 1.6. PROPOSITIONS FROM BOOK I: 27-48 78 * ID B Figure 1.6.7. [1.31] Take any point D on AB. Join CD [Postulate 1.1]. Choose any point E such that by joining points C and E we obtain ZDCE = ZADC [1.23]. By [1.29, Cor. 1], AB || CE. \u25a1 COROLLARY. 1. AB and CE in [1.31] may be replaced by rays and\/ or straight-line segments, mutatis mutandis. Definition: 43. The altitude of a triangle is the perpendicular segment from the trian- gle’s base to the base’s opposing vertex. Exercises. 1. Given the altitude of a triangle and the base angles, construct the trian- gle. 2. From a given point, construct a segment to a given segment such that the resultant angle is equal in measure to a given angle. Show that there are two solutions. 3. Prove the following construction for trisecting a given line AB: On AB, construct an equilateral AABC. Bisect the angles at points A, B by the lines AD, BD. Through D, construct parallels to AC, BC, intersecting AB at E, F. Claim: E and F are the points of trisection of AB. 4. Inscribe a square in a given equilateral triangle such that its base stands on a given side of the triangle. 5. Through two given points on two parallel lines, construct two segments forming a lozenge with given parallels. 6. Between two lines given in position, place a segment of given length which is parallel to a given line. Show that there are two solutions. Proposition 1.32. EXTERIOR ANGLES AND SUMS OF ANGLES IN A TRIANGLE. In any triangle, if one of the sides is extended, then: 1) the exterior angle equals the sum of the its interior and opposite angles; 2) the sum of the three interior angles of the triangle equals two right angles. 1.6. PROPOSITIONS FROM BOOK I: 27-48 79 PROOF. Construct AABC and wlog extend side AB to segment BD. We claim that the external angle ZCBD equals the sum of the two internal non- adjacent angles (ZBAC + ZACB) and that the sum of the three internal angles UBAC + ZACB + \/-ABC) equals two right angles. Or: ZCBD = ABAC + ZACB and ZBAC + ZACB + ZABC =two right angles. Figure 1.6.8. [1.32] Construct BE \\ AC [1.31]. Since BC intersects the parallels BE and AC, we have that ZEBC = ZACB [1.29]. Also, since AB intersects the parallels BE and AC, we have that ZEBD = ZBAC [1.29]. Since ZCBD = ZEBC + ZEBD, we have that ZCBD = ZACB + ZBAC This proves the first claim. Adding ZABC to this equality, we obtain ZABC + ZCBD = ZABC + ZAC\u00a3 + Z\u00a3AC But the sum ZA\u00a3?C + ZCBD equals two right angles [1.13]. Hence, the sum ZABC + ZACI? + ZBAC equals two right angles. \u25a1 COROLLARY. 1. If a right triangle is isosceles, then each base angle equals half of a right angle. COROLLARY. 2. If two triangles have two angles in one respectively equal to two angles in the other, then their remaining pair of angles is also equal. COROLLARY. 3. Since a quadrilateral can be divided into two triangles, the sum of its angles equals four right angles. 1.6. PROPOSITIONS FROM BOOK I: 27-48 80 COROLLARY. 4. If a figure of n sides is divided into triangles by drawing diagonals from any one of its angles, we will obtain (n-2) triangles. Hence, the sum of its angles equals 2 (n-2) right angles. COROLLARY. 5. If all the sides of any convex polygon are extended, then the sum of the external angles equals to four right angles. COROLLARY. 6. Each angle of an equilateral triangle equals two-thirds of a right angle. COROLLARY. 7. If one angle of a triangle equals the sum of the other two, then it is a right angle. COROLLARY. 8. Every right triangle can be divided into two isosceles tri- angles by a line constructed from the right angle to the hypotenuse. Exercises. 1. Trisect a right angle. 2. If the sides of a polygon of n sides are extended, then the sum of the angles between each alternate pair is equal to 2(n-4) right angles. 3. If the line which bisects the external vertical angle is parallel to the base, then the triangle is isosceles. 4. If two right triangles AABC, AABD are on the same hypotenuse AB and if the vertices C and D are joined, then the pair of angles standing opposite any side of the resulting quadrilateral are equal. 5. The three altitudes of a triangle are concurrent. Note: We are proving the existence of the orthocenteip^l of a triangle: the point where the three altitudes intersect, and one of a triangle’s points of concurrency^} (Hint: Solve using [1.34].) 6. The bisectors of two adjacent angles of a parallelogram are at right angles. (Hint: Solve using [1.34].) 7. The bisectors of the external angles of a quadrilateral form a circum- scribed quadrilateral, the sum of whose opposite angles equals two right an- gles. ^http : \/\/mathworld . wolfram . com\/Orthocenter . html 1.6. PROPOSITIONS FROM BOOK I: 27-48 81 8. If the three sides of one triangle are respectively perpendicular to those of another triangle, the triangles are equiangular. (This problem may be de- layed until the end of chapter 1.) 9. Construct a right triangle being given the hypotenuse and the sum or difference of the sides. 10. The angles made with the base of an isosceles triangle by altitudes from its endpoints on the equal sides are each equal to half the vertical angle. 11. The angle included between the internal bisector of one base angle of a triangle and the external bisector of the other base angle is equal to half the vertical angle. 12. In the construction of [1.18], prove that the angle ZDBC is equal to half the difference of the base angles. 13. If A, B, C denote the angles of a triangle, prove that {A+B), {B+C), and (A + C) are the angles of a triangle formed by any side, the bisectors of the external angles between that side, and the other extended sides. PROPOSITION 1.33. PARALLEL SEGMENTS. Segments which join ad- jacent endpoints of two equal, parallel segments are themselves parallel and equal in length. PROOF. If segments AC, BD join adjacent endpoints of two equal, parallel segments AB, CD, then AC = BD and AC || BD. Figure 1.6.9. [1.33] Join BC. Since AB || CD by hypothesis and BC intersects them, we have that ZABC = ZDCB [1.29]. Hence we have that AABC, ADCB have the sides AB, BC in one respectively equal to the sides DC, BC in the other where the interior angles ZABC, ZDCB are equal. By [1.4], AABC ^ ADCB, and so AC = BD and ZACB = ZCBD. Since ZACB, ZCBD are alternate angles, by [1.27], AC \\ BD. \u25a1 1.6. PROPOSITIONS FROM BOOK I: 27-48 82 COROLLARY. 1. [1.33] holds for straight lines and rays, mutatis mutandis. COROLLARY. 2. Figure BABDC is a parallelogram [Def. 1.39]. Exercises. 1. Prove that if two segments AB, BC are respectively equal and parallel to two other segments DE, EF, then the segment AC joining the endpoints of the former pair is equal in length to the segment DF joining the endpoints of the latter pair. Proposition 1.34. OPPOSITE SIDES AND OPPOSITE ANGLES OF PARALLELOGRAMS. The opposite sides and the opposite angles of a parallel- ogram are equal to one another and either diagonal bisects the parallelogram. PROOF. Construct BABCD. We claim that AB = CD, AC = BD, ZCAB = ZCDB, and ZACD = ZABD. Furthermore, we claim that either diagonal (CB, AD) bisects the parallelogram. Figure 1.6.10. [1.34] Join BC. Since AB || CD and BC intersects them, A ABC = ZDCB and ZACB = ZCBD [1.29]. Hence the triangles A ABC, ADCB have the two an- gles ZABC, ZACB in one respectively equal to the two angles ZBCD, ZCBD in the other with side BC in common. By [1.26], AABC ^ ADCB, and so AB = CD, AC = BD. We also have that ABAC = ZBDC, ZACD = ZACB+ZDCB and ZABD = ZCBD + ZABC. By the equalities above, we obtain ZACD = ZACB + ZDCB = ZCBD + ZABC = ZABD 1.6. PROPOSITIONS FROM BOOK I: 27-48 83 or, that opposite angles are equal. Since BABCD = AABC \u00a9 ADEF and AABC = ADCB (and hence the triangles have that same area), the diagonal bisects the parallelogram. The remaining case follows mutatis mutandis if we join AD instead of BC. \u25a1 COROLLARY. 1. The area of ‘BAB DC is double the area of A AC B (ABC D). COROLLARY. 2. If one angle of a parallelogram is a right angle, each of its angles are right angles. COROLLARY. 3. If two adjacent sides of a parallelogram are equal in length, then it is a lozenge. COROLLARY. 4. If both pairs of opposite sides of a quadrilateral are equal in length, it is a parallelogram. COROLLARY. 5. If both pairs of opposite angles of a quadrilateral are equal, it is a parallelogram. COROLLARY. 6. If the diagonals of a quadrilateral bisect each other, it is a parallelogram. COROLLARY. 7. If both diagonals of a quadrilateral bisect the quadrilat- eral, it is a parallelogram. COROLLARY. 8. If the adjacent sides of a parallelogram are equal, its di- agonals bisect its angles. COROLLARY. 9. If the adjacent sides of a parallelogram are equal, its di- agonals intersect at right angles. COROLLARY. 10. In a right parallelogram, the diagonals are equal in length. 1.6. PROPOSITIONS FROM BOOK I: 27-48 84 COROLLARY. 11. If the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a lozenge. COROLLARY. 12. If a diagonal of a parallelogram bisects the angles whose vertices it joins, the parallelogram is a lozenge. Exercises. 1. Show that the diagonals of a parallelogram bisect each other. 2. If the diagonals of a parallelogram are equal, each of its angles are right angles. 3. Divide a segment into any number of equal parts. 4. The segments joining the adjacent endpoints of two unequal parallel segments will meet when extended on the side of the shorter parallel. 5. If two opposite sides of a quadrilateral are parallel but unequal in length and the other pair are equal but not parallel, then its opposite angles are sup- plemental. 6. Construct a triangle being given the midpoints of its three sides. Proposition 1.35. AREAS OF PARALLELOGRAMS ON THE SAME BASE AND ON THE SAME PARALLELS. Parallelograms on the same base and between the same parallels are equal in area. PROOF. Parallelograms on the same base (BC) and between the same par- allels (AF, BC) are equal in area. The proof follows in three cases. ll Figure 1.6.11. [1.35], case 1 1. Construct the parallelograms BADCB, BFDBC on the common base BC. Notice that side AD of BADCB and side DF of BFDBC intersect only at point D. By [1.34], each parallelogram is double the area of the triangle ABCD. Hence, UADCB = BFDBC. 1.6. PROPOSITIONS FROM BOOK I: 27-48 85 A z E D F D E F 3 0 Figure 1.6.12. [1.35], cases 2 and 3 2. Construct the parallelograms BADCB and BEFCB such that side AD of BADCB and side FF of IEFCB intersect at more than one point (Fig. 1.6.12(a)). Because BABCD is a parallelogram, AD = BC [1.34]; because BBCEF is a parallelogram, \u00a3F = BC. Hence, AD = 2\u00a3F. Removing ED, we have that the sides BA, AE in ABAE are respectively equal to the two sides CD, DF in ACDF and that ZBAE = ZCDF [1.29, Cor. 1]. By [1.4], ABAE = ACDF. Notice that By the equality of the area of the triangles, it follows that BADCB = BEFCB. 3. Construct the parallelograms BADCB and BEFCB such that side AD of BADCB and side EF of BEFCB do not intersect (Fig. 1.6.12(\/3)). As in case 2, we have that AD = BC and EF = BC. Construct segment DE. Then we have that the sides BA, AE in ABAE are respectively equal to the two sides CD, DF in ACDF and that ZBAE = ZCDF [1.29]. By [1.4], ABAE = ACDF. As in case 2, it follows that BADCB = BEFCB. B Alternatively: PROOF. Construct the triangles ABAE, ACDF as well as the segment DE if necessary to create the quadrilateral AFCB. Notice that ABAE, ACDF have the sides AB, BE in one respectively equal to the sides DC, CF in the other [1.34] and that ZBAE = ZDCF [1.29, #8]. Hence, ABAE ^ ACDF. Since AFCB = BEFCB + ABAE = BADCB + ACDF AFCB = BEFCB + ABAE = BADCB + ACDF the proof follows. \u25a1 1.6. PROPOSITIONS FROM BOOK I: 27-48 86 Proposition 1.36. AREAS OF PARALLELOGRAMS ON EQUAL BASES AND ON THE SAME PARALLELS. Parallelograms on equal bases and on the same parallels are equal in area. PROOF. Parallelograms (BADCB, BEHGF) on equal bases (BC, FG) and standing between the same parallels {AH, BG) are equal in area. Join DE, CF, BE, CH. Since BEHGF is a parallelogram, FG = EH [1.34]. Since BG = FG by hypothesis, we have that BG = EH [Axiom 1.1]. Since BE, CH are also parallel and join the adjacent endpoints of EH, BG, BEBCH is a parallelogram. Again, since the parallelograms BADCB, BEBCH stand on the same base BG and between the same parallels BC, AH, RADCB = BEBCH [1.35]. Sim- ilarly, BEBCH = BEHGF. By [Axiom 1.1], RADCB = BEBCH. B Exercise. 1. Prove this proposition without joining BE, CH. PROPOSITION 1.37. TRIANGLES OF EQUAL AREA I. Triangles which stand on the same base and in the same parallels are equal in area. PROOF. We claim that triangles (AABC, ADBC) on the same base (BC) and standing between the same parallels (AD, BC) are equal in area. Construct BE \\ AC and CF \\ BD [1.31] and extend AD to segments AE and DF. It follows that the figures BAEBC, BDBCF are parallelograms. – Figure 1.6.13. [1.36] Figure 1.6.14. [1.37] 1.6. PROPOSITIONS FROM BOOK I: 27-48 87 Since they stand on the same base BC and between the same parallels BC, EF, BAEBC = BDBCF [1.35]. Notice that area of the triangle \u2022 AABC = BAEBC because the diagonal AB bisects UAEBC [1.34]. Similarly, \u25a0 ADBC = BDBCF. Since halves of equal magnitudes are equal [Axiom 1.7], we have that AABC = ADBC. \u25a1 Exercises. 1. If two triangles of equal area stand on the same base but on opposite sides, the segment joining their vertices is bisected by the base. 2. Construct a triangle equal in area to a given quadrilateral figure. 3. Construct a triangle equal in area to a given polygon. 4. Construct a lozenge equal in area to a given parallelogram and having a given side of the parallelogram for base. 5. Given the base and the area of a triangle, find the locus of the vertex. PROPOSITION 1.38. TRIANGLES OF EQUAL AREA II. Triangles which stand on equal bases and in the same parallels are equal in area. PROOF. By a construction analogous to [1.37], we have that the triangles are the halves of parallelograms, standing on equal bases and between the same parallels. Hence, they are the halves of equal parallelograms [1.36] and so are equal in area to each other. \u25a1 Exercises. 1. Every median of a triangle bisects the triangle. 2. If two triangles have two sides of one respectively equal to two sides of the other and where the interior angles are supplemental, their areas are equal. 3. If the base of a triangle is divided into any number of equal segments, then segments constructed from the vertex to the points of division divide the whole triangle into as many equal parts. 4. The diagonal of a parallelogram and segments from any point on the diagonal to the angular points through which the diagonal does not pass divide the parallelogram into four triangles which are equal (in a two by two fashion). 5. One diagonal of a quadrilateral bisects the other if and only if it also bisects the quadrilateral. 6. If two triangles AABC, AABD stand on the same base AB and between the same parallels, and if a parallel to AB meets the sides AC, BC at the points E, F as well as the sides AD, BD at the points G, H, then EF = GH. 1.6. PROPOSITIONS FROM BOOK I: 27-48 88 7. If instead of triangles on the same base we have triangles on equal bases and between the same parallels, the intercepts made by the sides of the triangles on any parallel to the bases are equal in length. 8. If the midpoints of any two sides of a triangle are joined, the triangle formed with the two half sides has an area equal to one-fourth of the whole. 9. The triangle whose vertices are the midpoints of two sides and any point in the base of another triangle has an area equal to one-fourth the area of that triangle. 10. Bisect a given triangle by a segment constructed from a given point in one of the sides. 11. Trisect a given triangle by three segments constructed from a given point within it. 12. Prove that any segment through the intersection of the diagonals of a parallelogram bisects the parallelogram. 13. The triangle formed by joining the midpoint of one of the non-parallel sides of a trapezium to the endpoints of the opposite side is equal in area to half the area of the trapezium. PROPOSITION 1.39. TRIANGLES OF EQUAL AREA III Triangles which are equal in area and stand on the same base and on the same side of the base also stand on the same parallels. PROOF. Equal triangles (ABAC, ABDC) on the same base (BC) and on the same side of the base also stand between the same parallels (AD, BC). Figure 1.6.15. [1.39] Join AD. If AD $ BC, suppose that AE \\ BC where the segments intersect at point E. Join EC. Since the triangles ABEC, ABAC stand on the same base BC and between the same parallels BC, AE, we have that ABEC = ABAC [1.37]. By hypothesis, ABAC = ABDC. Therefore, ABEC = ABDC [Axiom 1.1]. But ABDC = ABEC + AEDC, a contradiction. Hence, we must have that AD || BC. \u25a1 1.6. PROPOSITIONS FROM BOOK I: 27-48 89 PROPOSITION 1.40. TRIANGLES OF EQUAL AREA IV. Triangles which are equal in area and stand on equal bases and on the same side of their bases stand on the same parallels. PROOF. Triangles which are equal in area (AABC, ADEF) as well as stand on equal bases (BC, EF) and on the same side of their bases stand on the same parallels. Figure 1.6.16. [1.40] Join AD. If AD |f BF, let AG \\ BF. Join GF. Since the triangles AGEF, AABC stand on equal bases BC, EF and between the same parallels BF, AG, we have that AGEF = AABC [1.38]; but ADEF = AABC by hypothesis. Hence AGEF = ADEF [Axiom 1.1]. However, ADEF = AGEF + ADGF, a contradiction. Therefore, we must have that AD \\ BF. \u25a1 Exercises. 1. Triangles with equal bases and altitudes are equal in area. 2. The segment joining the midpoints of two sides of a triangle is parallel to the third because the medians from the endpoints of the base to these points will each bisect the original triangle. Hence, the two triangles whose base is the third side and whose vertices are the points of bisection are equal in area. 3. The parallel to any side of a triangle through the midpoint of another bisects the third. 4. The segments which connect the midpoints of the sides of a triangle divide it into four congruent triangles. 5. The segment which connects the midpoints of two sides of a triangle is equal in length to half the third side. 6. The midpoints of the four sides of a convex quadrilateral, taken in order, are the angular points of a parallelogram whose area is equal to half the area of the quadrilateral. 1.6. PROPOSITIONS FROM BOOK I: 27-48 90 7. The sum of the two parallel sides of a trapezium is double the length of the segment joining the midpoints of the two remaining sides. 8. The parallelogram formed by the segment which connects the midpoints of two sides of a triangle and any pair of parallels constructed through the same points to meet the third side is equal in area to half the area of the triangle. 9. The segment joining the midpoints of opposite sides of a quadrilateral and the segment joining the midpoints of its diagonals are concurrent. PROPOSITION 1.41. PARALLELOGRAMS AND TRIANGLES. If a paral- lelogram and a triangle stand on the same base and between the same parallels, then the parallelogram is double the area of the triangle. PROOF. If a parallelogram {BABCD) and a triangle (AEBC) stand on the same base (BC) and between the same parallels (AE, BC), then the parallelo- gram is double the area of the triangle. Figure 1.6.17. [1.41] Join AC. By [1.34], BABCD = 2 \u2022 A ABC, and by [1.37], AABC = AEBC. Therefore, BABCD = 2 \u2022 AEBC. B COROLLARY. 1. If a triangle and a parallelogram have equal altitudes and if the base of the triangle is double of the base of the parallelogram, their areas are equal. COROLLARY. 2. Suppose we have two triangles whose bases are two oppo- site sides of a parallelogram and which have any point between these sides as a common vertex. Then the sum of the areas of these triangles equals half the area of the parallelogram. 1.6. PROPOSITIONS FROM BOOK I: 27-48 91 Proposition 1.42. CONSTRUCTION OF PARALLELOGRAMS I. Given an arbitrary triangle and an arbitrary acute angle, it is possible to construct a parallelogram equal in area to the triangle which also contains the given angle. PROOF. We wish to construct a parallelogram equal in area to a given tri- angle (AABC) which contains an angle equal to a given angle (ZRDS). Figure 1.6.18. [1.42] Bisect AB at E and join EC. Construct ZBEF = ZRDS [1.23], CG \\ AB, and BG \\ EF [1.31]. We claim that BFEBG is the required parallelogram. Because AE = EB by construction, we have that AAEC = AEBC [1.38]. Therefore, AABC = 2- AEBC. We also have that BFEBG = 2- AEBC because each stands on the same base EB and between the same parallels EB and CG [1.41]. Therefore, BFEBG = AABC, and by construction, ZBEF = ZRDS. \u25a1 Proposition 1.43. COMPLEMENTARY AREAS OF PARALLELOGRAMS. Parallel segments through any point in one of the diagonals of a parallelogram divides the parallelogram into four smaller parallelograms: the two through which the diagonal does not pass are called the complements of the other two, and these complements are equal in area. PROOF. We claim that segments which are parallel to the sides of a par- allelogram BABCD (specifically EF,GH) and pass through any point (K) on one of the diagonals (AC) of BABCD divide EM.BCZ} into four smaller par- allelograms: the two through which the diagonal does not pass (BEBGK, BHKFD) are called the complements of the other two. We also claim that BEBGK = BHKFD. Because AC bisects the parallelograms BABCD, BAEKH, BKGCF, we have that AADC = AABC, AAHK = AAEK, and AKFC = AKGC [1.34]. Hence, we have that BEBGK = AABC – AAEK – AKGC ii BHKFD = AADC – AAHK – AKFC or simply BHKFD = BEBGK. B COROLLARY. 1. If through some point K within parallelogram BABCD we have constructed parallel segments to its sides in order to make the paral- lelograms BHKFD, BEBGK equal in area, then K is a point a the diagonal AC. COROLLARY. 2. BABGH = BAEFD and BEBCF = BHGCD. Proposition 1.44. CONSTRUCTION OF PARALLELOGRAMS II. Given an arbitrary triangle, an arbitrary angle, and an arbitrary segment, it is pos- sible to construct a parallelogram equal in area to the triangle which contains the given angle and has a side equal in length to the given segment. PROOF. On a given segment (AB), we wish to construct a parallelogram equal in area to a given triangle (ANPQ) which contains an equal to a given angle (ZRST). 1.6. PROPOSITIONS FROM BOOK I: 27-48 93 Figure 1.6.20. [1.44] Construct the parallelogram BBEFG such that BBEFG = ANPQ [1.42] where ZGBE = ZRST and where AB and BE form the segment AE. Also construct segment AH \\ BG [1.31]. Extend EG to intersect AH at point H. Join HB. Because HA \\ FE and HF intersects them, the sum ZAHF + ZHFE equals two right angles [1.29]. It follows that the sum ZBHF + ZHFE is less than two right angles since ZAHF = ZBHF + ZBHA. By [Axiom 1.12], if we extend segments HB and FE, they will intersect at some point K. Through K, construct KL || AB [1.31] and extend HA and GB to intersect KL at points L and M, respectively. We claim that BBALM is a parallelogram which fulfills the required conditions. By [1.43], BBALM = BFGBE. Recall that BFGBE = ANPQ by con- struction, and therefore BBALM = ANPQ. Again, ZABM = ZEBG [1.15], and ZEBG = ZRST by construction. Therefore, ZABM = ZRST. Finally, BBALM is constructed on the given segment AB. \u25a1 Proposition 1.45. CONSTRUCTION OF PARALLELOGRAMS III. Given an arbitrary angle and an arbitrary polygon, it is possible to construct a par- allelogram equal in area to the given polygon which contains an angle equal to the given angle. PROOF. We wish to construct a parallelogram equal in area to a given poly- gon (ABCD) which contains an angle equal to a given angle (ZLMN). 1.6. PROPOSITIONS FROM BOOK I: 27-48 94 Join BD. Construct a parallelogram BFEHG equal in area to the trian- gle AABD where ZFEH = ZLMN [1.42]. On segment GH, construct the parallelogram BGHKI such that BGHKI = ABCD where ZGHK = ZLMN [1.44]. We may continue to this algorithm for any additional triangles that re- main in ABCD. Upon exhaustion of this process, we claim that BFEKI is a parallelogram which fulfills the required conditions. Because ZLMN = ZFEH and ZGHK = ZLMN by construction, we have that ZGHK = ZFEH. From this, we obtain ZGHK + ZGHE = ZFEH + ZGHE But since HG || EF and EH intersects them, the sum ZFEH + AG HE equals two right angles [1.29]. Hence, the sum of ZGHK + ZGHE equals two right angles, and EH, HK form the segment EK [1.14, Cor. 1]. Similarly, because GH intersects the parallels FG, EK, the alternate an- gles ZFGH, ZGHK are equal [1.29]. From this, we obtain ZFGH + \/-HGI = ZGHK + Z-HGI Since GI \\ HK and GH intersects them, the sum ZGHK + ZHGI equals two right angles [1.29]. Hence, the sum ZFGH, +ZHGI equals two right an- gles, and FG and GI form the segment FI [1.14, Cor. 1]. Again, because [HFEHG and [UGHKI are parallelograms, EF and KI are each parallel to GH. By [1.30], we have that EF \\ KI and EK \\ FI. Therefore, IFEKI is a parallelogram. And because the parallelogram BFEHG = AABD by construction and BGHKI = ABCD, the parallelogram BFEKI = ABCD. Since ZFEH = ZLMN, the proof follows. \u25a1 Exercises. 1. Construct a rectangle equal to the sum of 2, 3, n number of polygons. 1.6. PROPOSITIONS FROM BOOK I: 27-48 95 2. Construct a rectangle equal in area to the difference in areas of two given figures. PROPOSITION 1.46. CONSTRUCTION OF A SQUARE I. Given an arbi- trary segment, it is possible to construct a square on that segment. PROOF. We wish to construct a square on a given segment (AB). A D 0 C Figure 1.6.22. [1.46] Construct AD _L AB [1.11] such that AD = AB [1.3]. Through point D, construct DC \\ AB [1.31], and through point B construct BC \\ AD where BC and DC intersect only at point C. We claim that BABCD is the required square. Because BAB CD is a parallelogram, AB = CD [1.34], and AB = AD by construction. Therefore, AD = CD and AD = BC [1.34] and the four sides of BABCD are equal. It follows that BABCD is a lozenge and ZDAB is a right angle. Therefore, AC is a square [Def. 1.30]. \u25a1 Exercises. 1. Two squares have equal side-lengths if and only if the squares are equal in area. 2. The parallelograms about the diagonal of a square are squares. 3. If on the four sides of a square, or on the sides which are extended, points are taken equidistant from the four angles, they will be the angular points of another square, and similarly for a regular pentagon, hexagon, etc. 4. Divide a given square into five equal parts: specifically, four right trian- gles and a square. PROPOSITION 1.47. THE PYTHAGOREAN THEOREM. In a right trian- gle, the square of the length of the side opposite the right angle (the hypotenuse) is equal to the sum of the squares of the remaining side-lengths. 1.6. PROPOSITIONS FROM BOOK I: 27-48 96 PROOF. In a right triangle (AABC), we claim that the square on the hy- potenuse (AB) is equal to the sum of the squares on the other two sides (AC, BC). Figure 1.6.23. [1.47] On the sides AB, BC, CA of AABC, construct squares [1.46]. Construct segment CL \\ AG. Join CG, BK. Because ZACB is right by hypothesis and ZACB. is right by construction, the sum ZACB + ZACB. equals two right an- gles. Therefore BC, CH form the segment BE [1.14]. Similarly, AC, CD for the segment AD. Because ZBAG, ZCAK are angles within a square, they are right angles. Hence, ZBAG = ZCAK; to each, add ZBAC, and we obtain ZCAG = ZKAB. Again, since BBAGF and BCHKA are squares, BA = AG, and CA = AK. Hence, the two triangles AC AG, AKAB have the sides CA, AG in one respec- tively equal to the sides KA, AB in the other such that their interior angles are equal (ZCAG = ZKAB). By [1.4], AC AG ^ AKAB. But BAGLO = 2 \u2022 AC AG because they are on the same base AG and between the same parallels (AG and CL), [1.41]. Similarly, the parallelogram BCHKA = 2 \u2022 AKAB because they stand on the same base AK and between the same parallels (AK and BH). Since doubles of equal magnitudes are equal [Axiom 1.6], the parallelo- gram BAGLO = BKACH. Similarly, it can be shown that the parallelogram 1.6. PROPOSITIONS FROM BOOK I: 27-48 97 BOLFB = BDCBE. Hence, BAGFB = BAGLO \u00a9 BOLFB = BKACH + BDCBE \u25a1 Alternatively: PROOF. Construct the squares as in Fig. 1.6.24. fl L P FIGURE 1.6.24. [1.47], alternate proof Join CG, BK, and through C construct OL || AG. Notice that ZGAK = ZGAC + + Z\u00a3Aif and that ZCAK are right angles. Removing ZBAC from each side of the equality, it follows that AC AG = ZBAK. Hence the triangles AC AG, ABAK have the sides CA = AK, AG = AB, and ZCAG = ZBAK; by [1.4], AC AG ^ ABAK. By [1.41], since the paral- lelograms BGAOL, BAKHC are respectively the doubles in area of the these triangles, we have that BGAOL = BAKHC. Similarly, BLOBF = BDEBC. Hence, . \u25a1 The alternative proof is shorter since it is not necessary to prove that AC, CD are in one segment. Similarly, the proposition may be proved by taking any of the eight figures formed by turning the squares in all possible directions. Another simplification of the proof can be obtained by considering that the point A is such that one of the triangles AC AG, ABAK can be turned round it in its own plane until it coincides with the other; hence, they are congruent. Exercises. 1.6. PROPOSITIONS FROM BOOK I: 27-48 98 1. The square on AC is equal in area to the rectangle BAB. AO, and the square on BBC = BAB. BO. (Note: BAB. AO denotes that rectangle formed by the segments AB and AO as well as the area of that rectangle.) 2. The square on BCO = BAO.OB. 3. Prove that AC 2 -BC 2 = A0 2 -B0 2 4. Find a segment whose square is equal to the sum of two given squares. 5. Given the base of a triangle and the difference of the squares of its sides, the locus of its vertex is a segment perpendicular to the base. 6. The transverse segments BK, CG are perpendicular to each other. 7. If EG is joined, then EG 2 = AC 2 + ABC 2 . 8. The square constructed on the sum of the sides of a right triangle ex- ceeds the square on the hypotenuse by four times the area of the triangle (see [1.46], Fig. 1.6.20, #3). More generally, if the vertical angle of a triangle is equal to the angle of a regular polygon of n sides, then the regular polygon of n sides, constructed on a segment equal to the sum of its sides exceeds the area of the regular polygon of n sides constructed on the base by n times the area of the triangle. 9. If AC and BK intersect at P and a segment is constructed through P which is parallel to BC, meeting AB at Q, then CP = PQ. 10. Each of the triangles AAGK and ABEF formed by joining adjacent corners of the squares is equal in area to the right triangle AABC. (Hint: use trigonometry.) 11. Find a segment whose square is equal to the difference of the squares on two segments. 12. The square on the difference of the sides AC, CB is less than the square on the hypotenuse by four times the area of the triangle. 13. If AE is joined, the segments AE, BK, CL, are concurrent. 14. In an equilateral triangle, three times the square on any side is equal to four times the square on the perpendicular to it from the opposite vertex. 15. We construct the square BBEFG on BE, a part of the side BC of a square BABCD, having its side BG in the continuation of AB. Divide the figure AGFECD into three parts which will form a square. 16. Four times the sum of the squares on the medians which bisect the sides of a right triangle is equal to five times the square on the hypotenuse. 17. If perpendiculars fall on the sides of a polygon from any point and if we divide each side into two segments, then the sum of the squares on one set of alternate segments is equal to the sum of the squares on the remaining set. 1.6. PROPOSITIONS FROM BOOK I: 27-48 99 18. The sum of the squares on segments constructed from any point to one pair of opposite angles of a rectangle is equal to the sum of the squares on the segments from the same point to the remaining pair. 19. Divide the hypotenuse of a right triangle into two parts such that the difference between their squares equals the square on one of the sides. 20. From the endpoints of the base of a triangle, let altitudes fall on the opposite sides. Prove that the sum of the rectangles contained by the sides and their lower segments is equal to the square on the base. Proposition 1.48. THE CONVERSE OF THE PYTHAGOREAN THEO- REM. If the square on one side of a triangle is equal to the sum of the squares on the remaining sides, then the angle opposite to the longest side (the hypotenuse) is a right angle. PROOF. If the square on one side (AB) of a triangle (AABC) equals the sum of the squares on the remaining sides (AC, CB), then the angle (ZACB) opposite to that side is a right angle. Or, if AB 2 = AC 2 + BC 2 , then ZACB is a right angle. Figure 1.6.25. [1.48] Construct CD _L CB [1.11] such that CD = CA [1.3]. Join BD. Because AC = CD, AC 2 = CD 2 . From this, we obtain AC 2 + CB 2 = CD 2 + CB 2 But AC 2 + CB 2 = AB 2 by hypothesis, and CD 2 + CB 2 = BD 2 [1.46]. It follows that AB 2 = BD 2 ; hence AB = BD [1.46, #1]. Again, because AC = CD by construction and CB is a common side to the triangles AACB, ADCB, we have that AB = DB and ZACB = ZDCB. But ZDCB is a right angle by construction, and so ZACB is a right angle. \u25a1 1.6. PROPOSITIONS FROM BOOK I: 27-48 100 The above proof forms an exception to Euclid’s demonstrations of converse propositions. The following is an indirect proof: FIGURE 1.6.26. [1.48], alternative proof PROOF. If CB \/ AC, construct CD _L CB such that CD = CB. Join AD. Then, as before, it can be shown that AD = AB. This is contrary to [1.7]. Hence, ZACB is a right angle. \u25a1 Examination questions on chapter 1. 1. What is geometry? 2. What is geometric magnitude? 3. Name the primary concepts of geometry. (Ans. Points, lines, surfaces, and solids.) 4. What kinds of lines exist in geometry (Ans. Straight and curved.) 5. How is a straight line generated? (Ans. By connecting any three collinear points.) 6. How is a curved line generated? (Ans. By connecting any three non- collinear points.) 7. How may surfaces be divided? (Ans. Into planes and curved surfaces.) 8. How may a plane surface be generated? 9. Why has a point no dimensions? 10. Does a line have either width nor thickness? 11. How many dimensions does a surface possess? 12. What is plane geometry? 13. What portion of plane geometry forms the subject of this chapter? 14. What is the subject-matter of the remaining chapters? 15. How is a proposition proved indirectly? 16. What is meant by the inverse of a proposition? 17. What proposition is an instance of the Rule of Symmetry? 18. What are congruent figures? 19. What is another way to describe congruent figures? (Ans. They are said to be identically equal.) 1.6. PROPOSITIONS FROM BOOK I: 27-48 101 20. Mention all the instances of equality which are not congruence that occur in chapter 1. 21. What is the difference between the symbols denoting congruence and identity? 22. Define adjacent, exterior, interior, and alternate angles, respectively. 23. What is meant by the projection of one line on another? 24. What are meant by the medians of a triangle? 25. What is meant by the third diagonal of a quadrilateral? 26. Mention some propositions in chapter 1 which are particular cases of more general ones that follow. 27. What is the sum of all the exterior angles of any polygon equal to? 28. How many conditions must be given in order to construct a triangle? (Ans. Three; such as the three sides, or two sides and an angle, etc.) Chapter 1 exercises. 1. Any triangle is equal to a fourth part of the area which is formed by constructing through each vertex a line which is parallel to its opposite side. 2. The three altitudes of the first triangle in #1 are the altitudes at the midpoints of the sides of the second triangle. 3. Through a given point, construct a line so that the portion intercepted by the segments of a given angle are bisected at the point. 4. The three medians of a triangle are concurrent. (Note: we are proving the existence of the centroid of a triangle.) 5. Construct a triangle given two sides and the median of the third side. 6. In every triangle, the sum of the medians is less than the perimeter but greater than three-fourths of the perimeter. 7. Construct a triangle given a side and the two medians of the remaining sides. 8. Construct a triangle given the three medians. 9. The angle included between the perpendicular from the vertical angle of a triangle on the base and the bisector of the vertical angle is equal to half the difference of the base angles. 10. Find in two parallels two points which are equidistant from a given point and whose connecting line is parallel to a given line. 11. Construct a parallelogram given two diagonals and a side. 12. The shortest median of a triangle corresponds to the largest side. 13. Find in two parallels two points standing opposite a right angle at a given point and which are equally distant from it. 1.6. PROPOSITIONS FROM BOOK I: 27-48 102 14. The sum of the distances of any point in the base of an isosceles triangle from the equal sides is equal to the distance of either endpoint of the base from the opposite side. 15. The three perpendiculars at the midpoints of the sides of a triangle are concurrent. Hence, prove that perpendiculars from the vertices on the opposite sides are concurrent [see #2]. 16. Inscribe a lozenge in a triangle having for an angle one angle of the triangle. 17. Inscribe a square in a triangle having its base on a side of the triangle. 18. Find the locus of a point, the sum or the difference of whose distance from two fixed lines is equal to a given length. 19. The sum of the perpendiculars from any point in the interior of an equi- lateral triangle is equal to the perpendicular from any vertex on the opposite side. 20. Find a point in one of the sides of a triangle such that the sum of the intercepts made by the other sides on parallels constructed from the same point to these sides are equal to a given length. 21. If two angles exist such that their segments are respectively parallel, then their bisectors are either parallel or perpendicular. 22. Inscribe in a given triangle a parallelogram whose diagonals intersect at a given point. 23. Construct a quadrilateral where the four sides and the position of the midpoints of two opposite sides are given. 24. The bases of two or more triangles having a common vertex are given, both in magnitude and position, and the sum of the areas is given. Prove that the locus of the vertex is a straight line. 25. If the sum of the perpendiculars from a given point on the sides of a given polygon is given, then the locus of the point is a straight line. 26. If AABC is an isosceles triangle whose equal sides are AB, AC and if B’C is any secant cutting the equal sides at B’ , C , such that AB’ + AC = AB + AC, prove that B’C > BC. 27. If A, B are two given points and P is a point on a given line L, prove that the difference between AP and PB is a maximum when L bisects the angle ZAPB. Show that their sum is a minimum if it bisects the supplement. 28. Bisect a quadrilateral by a segment constructed from one of its angular points. 29. If AD and BC are two parallel lines cut obliquely by AB and perpen- dicularly by AC, and between these lines we construct BED, cutting AC at point E such that ED = 2AB, prove that the angle ZDBC = \u00a7 \u2022 ZABC. 1.6. PROPOSITIONS FROM BOOK I: 27-48 103 30. If O is the point of concurrence of the bisectors of the angles of the tri- angle AABC, if AO is extended to intersect BC at D, and if OE is constructed from O such that OE _L BC, prove that the ZBOD = ZCOE. 31. The angle made by the bisectors of two consecutive angles of a convex quadrilateral is equal to half the sum of the remaining angles; the angle made by the bisectors of two opposite angles is equal to half the difference of the two other angles. 32. If in the construction of [1.47] we join EF, KG, then EF 2 + KG 2 = 5AB 2 . 33. Given the midpoints of the sides of a convex polygon of an odd number of sides, construct the polygon. 34. Trisect a quadrilateral by lines constructed from one of its angles. 35. Given the base of a triangle in magnitude and position and the sum of the sides, prove that the perpendicular at either endpoint of the base to the adjacent side and the external bisector of the vertical angle meet on a given line perpendicular to the base. 36. The bisectors of the angles of a convex quadrilateral form a quadri- lateral whose opposite angles are supplemental. If the first quadrilateral is a parallelogram, the second is a rectangle; if the first is a rectangle, the second is a square. 37. Suppose that the midpoints of the sides AB, BC, CA of a triangle are respectively D, E, F and that DG \\ BF and intersects EF. Prove that the sides of the triangle ADCG are respectively equal to the three medians of the triangle AABC. 38. Find the path of a pool ball started from a given point which, after being reflected from the four sides of the table, will pass through another given point. (Assume that the ball does not enter a pocket.) 39. If two lines bisecting two angles of a triangle and terminated by the opposite sides are equal, prove that the triangle is isosceles. 40. State and prove the proposition corresponding to #37 when the base and difference of the sides are given. 41. If a square is inscribed in a triangle, the rectangle under its side and the sum of the base and altitude is equal to twice the area of the triangle. 42. If AB, AC are equal sides of an isosceles triangle and if BD _L AC, prove that BC 2 = 2AC.CD. 43. Given the base of a triangle, the difference of the base angles, and the sum or difference of the sides, construct it. 44. Given the base of a triangle, the median that bisects the base, and the area, construct it. 1.6. PROPOSITIONS FROM BOOK I: 27-48 104 45. If the diagonals AC, BD of a quadrilateral ABCD intersect at E and are bisected at the points F, G, then 4 \u2022 AEFG = (AEB + ECD)-(AED + EBC) 46. If squares are constructed on the sides of any triangle, the lines of connection of the adjacent corners are respectively: (a) the doubles of the medians of the triangle; (b) perpendicular to them. CHAPTER 2 Rectangles This chapter proves a number of propositions which demonstrate elemen- tary algebraic statements that are more familiar to us in the form of equa- tions. Algebra as we know it had not been developed when Euclid wrote “The Elements”. Hence, the results are more of historical importance than practical use except when they appear in subsequent proofs. As such, Book II appears here in truncated form. Students should feel free to solve the exercises in this chapter algebraically. Note that Axioms and Mathematical Properties from chapter 1 will not generally be cited. 2.1. Definitions 1. If a point C is taken on a segment AB, point C is the point of division between segments AC and CB. 2. If the segment AB is extended to point C, then point C is called a point of external division. a c F 4 * * FIGURE 2.1.1. [Def. 2.1] above, [Def 2.2] below 3. A parallelogram whose angles are right angles is called a rectangle. 105 2.2. AXIOMS 106 i.: :. L! FIGURE 2.1.2. [Def. 2.3 and 2.4] 4. A rectangle is said to be contained by any two adjacent sides. Thus, the rectangle ABCD is said to be contained by AB, AD, or by AB, BC, etc. 5. The rectangle contained by two separate segments (such as AB and CD in Fig 2.1.2) is the parallelogram formed by constructing a perpendicular to AB at A which is equal in length to CD and constructing parallels. The area of the rectangle may also be denoted as AB.CD. 6. In any parallelogram, a figure which is composed of either of the paral- lelograms about a diagonal and the two complements is called a gnomon [see 1.43]. Thus, if we remove either of the parallelograms BAGDE, BOFCH from the parallelogram [HADCB, the remainder is a gnomon. 2.2. Axioms 1. A semicircle (half-circle) may be constructed given only its center point and a radius. : Figure 2.1.3. [Def. 2.6] 2.3. PROPOSITIONS FROM BOOK II 107 2.3. Propositions from Book II PROPOSITION 2.1. Suppose that two segments (AB, BD) which intersect at one and only one point (B) are constructed such that one segment (BD) is divided into an arbitrary but finite number of segments (BC, CE, EF, FD). Then the rectangle contained by the two segments AB and BD is equal in area to the sum of the areas of the rectangles contained by AB and the subsegments of the divided segment. COROLLARY. 1. Algebraically, [2.1] states that the area AB \u25a0 BD = AB \u25a0 BC + AB \u25a0 CE + AB \u25a0 EF + AB \u25a0 FD More generally, it states that if y = VI+V2 + \u2022\u2022\u2022 J ryn> then xy = xyi+xy 2 + …+xy n . COROLLARY. 2. The rectangle contained by a segment and the difference of two other segments equals the difference of the rectangles contained by the segment and each of the others. COROLLARY. 3. The area of a triangle is equal to half the rectangle con- tained by its base and perpendicular. PROPOSITION 2.2. If a segment (AB) is divided into any two subsegments at a point (C), then the square on the segment is equal to the sum of the rectan- gles contained by the whole and each of the subsegments (AC, CB). A |G ^ ‘ \u2014 \u20144\u2014 \u2014 \u2014 iL H_ Figure 2.3.1. [2.1] 2.3. PROPOSITIONS FROM BOOK II 108 F 0 Figure 2.3.2. [2.2] COROLLARY. 1. Algebraically, [2.2] is a special case of [2.1] when n = 2. Specifically, it states that AF \u2022 FD = AF \u2022 FE + AF \u2022 ED orify = yi+ y 2 , then xy = xyi + xy 2 . PROPOSITION 2.3. If a segment (AB) is divided into two subsegments (at C), the rectangle contained by the whole line and either subsegment (CB or CF) is equal to the square on that segment together with the rectangle contained by each of the segments. Figure 2.3.3. [2.3] COROLLARY. 1. Algebraically, [2.3] states that ifx = y+z, then xy = y 2 +yz. PROPOSITION 2.4. If a segment (AB) is divided into any two parts (at C), the square on the whole segment is equal to the sum of the squares on the sub- segments (AC, CB) together with twice the area of their rectangle. 2.3. PROPOSITIONS FROM BOOK II 109 E Figure 2.3.4. [2.4] COROLLARY. 1. Algebraically, [2.4] states that if x = y + z, then x 2 = y 2 + 2yz + z 2 . COROLLARY. 2. The parallelograms about the diagonal of a square are squares. COROLLARY. 3. The square on a segment is equal in area to four times the square on its half. COROLLARY. 4. If a segment is divided into any number of sub segments, the square on the whole is equal to the sum of the squares on all the subseg- ments, together with twice the sum of the rectangles contained by the several distinct pairs of sub segments. Exercises. 1. Prove [2.4] by using [2.2] and [2.3]. 2. If from the vertical angle of a right triangle a perpendicular falls on the hypotenuse, its square equals the area of the rectangle contained by the segments of the hypotenuse. 3. If from the hypotenuse of a right triangle subsegments are cut off equal to the adjacent sides, prove that the square on the middle segment is equal in area to twice the rectangle contained by the segments at either end. 4. In any right triangle, the square on the sum of the hypotenuse and perpendicular from the right angle on the hypotenuse exceeds the square on the sum of the sides by the square on the perpendicular. 5. The square on the perimeter of a right-angled triangle equals twice the rectangle contained by the sum of the hypotenuse and one side and the sum of the hypotenuse and the other side. 2.3. PROPOSITIONS FROM BOOK II 110 PROPOSITION 2.5. If a segment (AB) is divided into two equal parts (at C) and also into two unequal parts (at D), the rectangle (AD, DB) contained by the unequal parts together with the square on the part between the points of section () is equal to the square on half the line. + \u2014 t E K. e- i |_ ^ A Figure 2.3.5. [2.5] COROLLARY. 1. Algebraically, [2.5] states that _ (x + y) 2 (x – y) 2 Xy ~ 2 + 2 This may also be expressed as AD.DB + CD 2 = CB 2 = CA 2 . COROLLARY. 2. The rectangle AD.DB is the rectangle contained by the sum of the segments AC, CD and their difference, and we have proved it equal to the difference between the square on AC and the square on CD. Hence the difference of the squares on two segments is equal to the rectangle contained by their sum and their difference. COROLLARY. 3. The perimeter of the rectangle AH = 2AB, and is therefore independent of the position of the point D on the line AB. The area of the same rectangle is less than the square on half the segment by the square on the subsegment between D and the midpoint of the line; therefore, when D is the midpoint, the rectangle will have the maximum area. Hence, of all rectangles having the same perimeter, the square has the greatest area. Exercises. 1. Divide a given segment so that the rectangle contained by its parts has a maximum area. 2. Divide a given segment so that the rectangle contained by its subseg- ments is equal to a given square, not exceeding the square on half the given line. 2.3. PROPOSITIONS FROM BOOK II 111 3. The rectangle contained by the sum and the difference of two sides of a triangle is equal to the rectangle contained by the base and the difference of the segments of the base made by the perpendicular from the vertex. 4. The difference of the sides of a triangle is less than the difference of the segments of the base made by the perpendicular from the vertex. 5. The difference between the square on one of the equal sides of an isosce- les triangle and the square on any segment constructed from the vertex to a point in the base is equal to the rectangle contained by the segments of the base. 6. The square on either side of a right triangle is equal to the rectangle contained by the sum and the difference of the hypotenuse and the other side. PROPOSITION 2.6. If a segment (AB) is bisected (at C) and extended to a segment (BD), the rectangle contained by the segments (AD, BD) made by the endpoint of the second segment ( D) together with the square on half of the segment (CB) equals the square on the segment between the midpoint and the endpoint of the second segment. E G F Figure 2.3.6. [2.6] COROLLARY. 1. Algebraically, [2.6] states that x(x-b) = (x- b -) 2 -( b -) 2 This may also be expressed as AD.BD + CB 2 = CD 2 . Exercises. 1. Show that [2.6] is reduced to [2.5] by extending the line in the opposite direction. 2. Divide a given segment externally so that the rectangle contained by its subsegments is equal to the square on a given line. 2.3. PROPOSITIONS FROM BOOK II 112 3. Given the difference of two segments and the rectangle contained by them, find the subsegments. 4. The rectangle contained by any two segments equals the square on half the sum minus the square on half the difference. 5. Given the sum or the difference of two lines and the difference of their squares, find the lines. 6. If from the vertex C of an isosceles triangle a segment CD is constructed to any point in the extended base, prove that CD 2 -CB 2 = AD.DB. 7. Give a common statement which will include [2.5] and [2.6]. PROPOSITION 2.7. If a segment (AB) is divided into any two parts (at C), the sum of the squares on the whole segment (AB) and either subsegment (CB) equals twice the rectangle (double AB, CB) contained by the whole segment and that subsegment, together with the square on the remaining segment. A C B Figure 2.3.7. [2.7] COROLLARY. 1. Algebraically, [2.7] states that ifx = y + z, then x 2 + z 2 = 2xz + y 2 ; equivalently, this can be stated as (x \u2014 z) 2 = y 2 . Or, AB 2 + BC 2 = 2 \u2022 AB \u25a0 BC + AC 2 COROLLARY. 2. Comparison of [2.4] and [2.7]: [2.4]: square on sum = sum of squares + twice rectangle [2.7]: square on difference = sum of squares-twice rectangle PROPOSITION 2.8. If a segment (AB) is divided into two parts (at C), the square on the sum of the whole segment (AB) and either subsegment (BC) equals four times the rectangle contained by the whole line (AB) and that seg- ment, together with the square on the other segment (AC). COROLLARY. 1. Algebraically, [2.8] states that if x = y + z, then (x + y) 2 = 4xy + z 2 = 4xy + (x \u2014 y) 2 Exercises. 1. In [1.47], if EF, GK are joined, prove that EF 2 -C0 2 = (AB + BO) 2 . 2. Prove that GK 2 -EF 2 = 3AB \u2022 (AO-BO). 3. Given that the difference of two segments equals R and the area of their rectangle equals 4R 2 , find the segments. PROPOSITION 2.9. If a segment (AB) is bisected (at C) and divided into two unequal segments (at D), the sum of the squares on the unequal subsegments (AD, DB) is double the sum of the squares on half the line (AC) and on the segment (CD) between the points of section. A a Figure 2.3.9. [2.9] 2.3. PROPOSITIONS FROM BOOK II 114 COROLLARY. Algebraically, [2.9] states that (y + z) 2 + (y – z) 2 = 2(y + z) 2 . Exercises. 1. The sum of the squares on the subsegments of a larger segment of fixed length is a minimum when it is bisected. 2. Divide a given segment internally so that the sum of the squares on the subsegments equals a given square and state the limitation to its possibility. 3. If a segment AB is bisected at C and divided unequally in D, then AD 2 + DB 2 = 2AD.DB + 4CD 2 . 4. Twice the square on the segment joining any point in the hypotenuse of a right isosceles triangle to the vertex is equal to the sum of the squares on the segments of the hypotenuse. 5. If a segment is divided into any number of subsegments, the contin- ued product of all the parts is a maximum and the sum of their squares is a minimum when all the parts are equal. PROPOSITION 2.10. If a segment (AB) is bisected (at C) and is extended to a segment (BD), the sum of the squares on the segments (AD, DB) made by the endpoint ( D) is equal to twice the square on half the line and twice the square on the segment between the points of that section. COROLLARY. 1. Algebraically, [2.10] states the same result as Proposition F Figure 2.3.10. [2.10] 2.9:(y + z) 2 + (y-z) 2 = 2(y + z) 2 . COROLLARY. 2. The square on the sum of any two segments plus the square on their difference equals twice the sum of their squares. 2.3. PROPOSITIONS FROM BOOK II 115 COROLLARY. 3. The sum of the squares on any two segments is equal to twice the square on half the sum plus twice the square on half the difference of the lines. COROLLARY. 4. If a segment is cut into two unequal subsegments and also into two equal subsegments, the sum of the squares on the two unequal subsegments exceeds the sum of the squares on the two equal subsegments by the sum of the squares of the two differences between the equal and unequal subsegments. Exercises. 1. Given the sum or the difference of any two segments and the sum of their squares, find the segments. 2. The sum of the squares on two sides AC, CB of a triangle is equal to twice the square on half the base AB and twice the square on the median which bisects AB. 3. If the base of a triangle is given both in magnitude and position and the sum of the squares on the sides in magnitude, the locus of the vertex is a circle. 4. If in AABC a point D on the base BC exists such that BA 2 + BD 2 = CA 2 + CD 2 , prove that the midpoint of AD is equally distant from both B and C. PROPOSITION 2.11. It is possible to divide a given segment (AB) into two segments (at H) such that the rectangle (AB, BH) contained by the whole line and one segment is equal in area to the square on the other segment (AH). Figure 2.3.11. [2.11] Definition: A segment divided as in this proposition is said to be divided in “extreme and mean ratio.” 2.3. PROPOSITIONS FROM BOOK II 116 COROLLARY. 1. Algebraically, [2.11] solves the equation AB \u2022 BH = AH 2 , or a(a \u2014 x) = x 2 . Specifically, 2 a (a \u2014 x) = x a 2 \u2014 ax = x 2 .2 x = -\u00a7(l\u00b1>\/5) Note that 7 = 1+ 2 V ^ is called the Golden Ratio^ COROLLARY. 2. The segment CF is divided in “extreme and mean ratio” at A. COROLLARY. 3. If from the greater segment CA of CF we take a segment equal to AF, it is evident that CA will be divided into parts respectively equal to AH, HB. Hence, if a segment is divided in extreme and mean ratio, the greater segment will be cut in the same manner by taking on it a part equal to the less, and the less will be similarly divided by taking on it a part equal to the difference, and so on. AD EC b \u25a0 * \u25a0 \u25a0 i Figure 2.3.12. [2.11], Cor. 4 COROLLARY. 4. Let AB be divided in “extreme and mean ratio” at C. It is evident ([2.11], Cor. 2) that AC > CB. Cut off CD = CB. Then by ([2.11], Cor. 2), AC is cut in “extreme and mean ratio” at D, and CD > AD. Next, cut off DE = AD, and in the same manner we have DE > EC, and so on. Since CD > AD, it is evident that CD is not a common measure of AC and CB, and therefore not a common measure of AB and AC. Similarly, AD is not a common measure of AC and CD and so is therefore not a common measure of AB and AC. Hence, no matter how far we proceed, we cannot arrive at any remainder which will be a common measure of AB and AC. Hence, the parts of a line divided in “extreme and mean ratio” are incommensurable (i.e., their ratio will never be a rational number). https : \/\/en.wikipedia. org\/wiki\/Golden_ratio 2.3. PROPOSITIONS FROM BOOK II 117 See also [6.30] where we divide a given segment (AB) into its “extreme and mean ratio”; that is, we divide a line segment AB at point C such that AB \u25a0 BC = AC 2 . Exercises. 1. The difference between the squares on the segments of a line divided in “extreme and mean ratio” is equal to their rectangle. 2. In a right triangle, if the square on one side is equal to the rectangle con- tained by the hypotenuse and the other side, the hypotenuse is cut in “extreme and mean ratio” by the perpendicular on it from the right angle. 3. If AB is cut in “extreme and mean ratio” at C, prove that (a) AB 2 + BC 2 = 3AC 2 (b) (AB + BC) 2 = 5AC 2 4. The three lines joining the pairs of points G, B; F, D; A, K, in the construction of [2.11] are parallel. 5. If CH intersects BE at O, AO _L CH. 6. If CH is extended, then CH _L BF. 7. Suppose that AABC is a right-angled triangle having AB = 2AC. If AH is equal to the difference between BC and AC, then AB is divided in “extreme and mean ratio” at H. PROPOSITION 2.12. On an obtuse-angled triangle (AABC), the square on the side opposite the obtuse angle (AB) exceeds the sum of the squares on the sides (BC, CA) containing the obtuse angle by twice the rectangle contained by either of them (BC) and its extension (CD) to meet a perpendicular (AD) on it from the opposite angle. Figure 2.3.13. [2.12] COROLLARY. 1. Algebraically, [2.12] states that in an obtuse triangle AB 2 = AC 2 + BC 2 + 2 \u2022 BC \u2022 CD. This is extremely close to stating the law of cosines : c 2 = a 2 + b 2 -2ab \u2022 cos(a). 2.3. PROPOSITIONS FROM BOOK II 118 COROLLARY. 2. If perpendiculars from A and B to the opposite sides meet them in H and D, the rectangle AC.CH is equal in area to the rectangle BC.CD (or BAC.CH = BBC.CD). Exercises. 1. If the angle ZACB of a triangle is equal to twice the angle of an equilat- eral triangle, then AB 2 = BC 2 + CA 2 + BC.CA. 2. Suppose that ABCD is a quadrilateral whose opposite angles at points B and D are right, and when AD, BC are extended meet at E, prove that AE.DE = BE.CE. 3. If A ABC is a right triangle and BD is a perpendicular on the hy- potenuse AC, prove that AB.DC = BD.BC. 4. If a segment AB is divided at C so that AC 2 = 2CB 2 , prove that AB 2 + BC 2 = 2AB.AC. 5. If AB is the diameter of a semicircle, find a point C in AB such that, joining C to a fixed point D in the circumference and constructing a perpen- dicular CE intersecting the circumference at E, then CE 2 -CD 2 is equal to a given square. 6. If the square of a segment CD, constructed from the angle C of an equilateral triangle AABC to a point D on the extended side AB is equal to 2AB 2 , prove that AD is cut in “extreme and mean ratio” at B. PROPOSITION 2.13. In any triangle (AABC), the square on any side op- posite an acute angle (at C) is less than the sum of the squares on the sides containing that angle by twice the rectangle (BC, CD) contained by either of them (BC) and the intercept (CD) between the acute angle and the foot of the perpendicular on it from the opposite angle. Figure 2.3.14. [2.13] COROLLARY. 1. Algebraically, [2.13] states the same result as [2.12]. 2.3. PROPOSITIONS FROM BOOK II 119 Exercises. 1. If the angle at point C of the AACB is equal to an angle of an equilateral triangle, then AB 2 = AC 2 + BC 2 -AC.BC. 2. The sum of the squares on the diagonals of a quadrilateral, together with four times the square on the line joining their midpoints, is equal to the sum of the squares on its sides. 3. Find a point C in a given extended segment AB such that AC 2 + BC 2 = 2AC.BC. Proposition 2.14. CONSTRUCTION OF A SQUARE II. It is possible to construct a square equal to a given an arbitrary polygon. PROOF. We wish to construct a square equal in area to a given polygon (MNPQ). Figure 2.3.15. [2.14] Construct the rectangle BABCD equal in area to MNPQ [1.45]. If the adjacent sides of [HABCD (AB, BC) are equal, HABCD is a square and the proof follows. Otherwise, extend AB to E such that BE = BC. Bisect AE at F, and with F as center and FE as radius, construct the semicircle AGE. Extend CB to the semicircle at G. We claim that the square constructed on BG is equal in area to MNPQ. To see this, join FG. Because AE is divided equally at F and unequally at B, AB.BE + FB 2 = FE 2 = FG 2 [2.5]. But FG 2 = FB 2 + BG 2 by [1.47]. Therefore, the rectangle AB.BE + FB 2 = FB 2 + BG 2 . 2.3. PROPOSITIONS FROM BOOK II 120 Subtracting FB 2 from both sides of the equality, we have that the rectangle AB.BE = BG 2 . Since BE = BC, the rectangle AB.BE = BABCD. Therefore BG 2 = BABCD which is equal in area to the given polygon, MNPQ. \u25a1 COROLLARY. 1. The square on the perpendicular from any point on a semi- circle to the diameter is equal to the rectangle contained by the segments of the diameter Exercises. 1. Given the difference of the squares on two segments and their rectangle, find the segments. Examination questions on chapter 2. 1. What is the subject-matter of chapter 2? (Ans. Theory of rectangles.) 2. What is a rectangle? A gnomon? 3. What is a square inch? A square foot? A square mile? (Ans. The square constructed on a line whose length is an inch, a foot, or a mile.) 4. When is a line said to be divided internally? When externally? 5. How is the area of a rectangle determined? 6. How is a line divided so that the rectangle contained by its segments is a maximum? 7. How is the area of a parallelogram found? 8. What is the altitude of a parallelogram whose base is 65 meters and area 1430 square meters? 9. How is a segment divided when the sum of the squares on its subseg- ments is a minimum? 10. The area of a rectangle is 108.60 square meters and its perimeter is 48.20 linear meters. Find its dimensions. 11. What proposition in chapter 2 expresses the distributive law of multi- plication? 12. On what proposition is the rule for extracting the square root founded? 13. Compare [1.47], [2.12], and [2.13]. 14. If the sides of a triangle are expressed algebraically by x 2 + 1, x 2 -l, and 2x units, respectively, prove that it is a right triangle. 15. How would you construct a square whose area would be exactly an acre? Give a solution using [1.47]. 16. What is meant by incommensurable lines? Give an example from chap- ter 2. 2.3. PROPOSITIONS FROM BOOK II 121 17. Prove that a side and the diagonal of a square are incommensurable. 18. The diagonals of a lozenge are 16 and 30 meters respectively. Find the length of a side. 19. The diagonal of a rectangle is 4.25 inches, and its area is 7.50 square inches. What are its dimensions? 20. The three sides of a triangle are 8, 11, 15. Prove that it has an obtuse angle. 21. The sides of a triangle are 13, 14, 15. Find the lengths of its medians. Also find the lengths of its perpendiculars and prove that all its angles are acute. 22. If the sides of a triangle are expressed by m 2 + n 2 , m 2 -n 2 , and 2mn linear units, respectively, prove that it is right-angled. Chapter 2 exercises. 1. The squares on the diagonals of a quadrilateral are together double the sum of the squares on the segments joining the midpoints of opposite sides. 2. If the medians of a triangle intersect at O, then AB 2 + BC 2 + CA 2 = 3(0 A 2 + OB 2 -f OC 2 ). 3. Through a given point O, construct three segments OA, OB, OC of given lengths such that their endpoints are collinear and that AB = BC. 4. If in any quadrilateral two opposite sides are bisected, the sum of the squares on the other two sides, together with the sum of the squares on the diagonals, is equal to the sum of the squares on the bisected sides together with four times the square on the line joining the points of bisection. 5. If squares are constructed on the sides of any triangle, the sum of the squares on the segments joining the adjacent corners is equal to three times the sum of the squares on the sides of the triangle. 6. Divide a given segment into two parts so that the rectangle contained by the whole and one segment is equal to any multiple of the square on the other segment. 7. If P is any point in the diameter AB of a semicircle and CD is any parallel chord, then CP 2 + PD 2 = AP 2 + PB 2 . 8. If A, B, C, D are four collinear points taken in order, then AB.CD + BC.AD = AC.BD. 9. Three times the sum of the squares on the sides of any pentagon exceeds the sum of the squares on its diagonals by four times the sum of the squares on the segments joining the midpoints of the diagonals. 10. In any triangle, three times the sum of the squares on the sides is equal to four times the sum of the squares on the medians. 2.3. PROPOSITIONS FROM BOOK II 122 11. If perpendiculars are constructed from the angular points of a square to any line, the sum of the squares on the perpendiculars from one pair of opposite angles exceeds twice the rectangle of the perpendiculars from the other pair by the area of the square. 12. If the base AB of a triangle is divided at D such that mAD = nBD, then mAC 2 + nBC 2 = mAD 2 + nDB 2 + (m + n)CD 2 . 13. If the point D is taken on the extended segment AB such that mAD = nDB, then mAC 2 -nBC 2 = mAD 2 -nDB 2 + (m-n)CD 2 . 14. Given the base of a triangle in magnitude and position as well as the sum or the difference of m times the square on one side and n times the square on the other side in magnitude, then the locus of the vertex is a circle. 15. Any rectangle is equal in area to half the rectangle contained by the diagonals of squares constructed on its adjacent sides. 16. If A, B, C, … are any finite number of fixed points and P a movable point, find the locus of P HAP 2 + BP 2 + CP 2 + … is given. 17. If the area of a rectangle is given, its perimeter is a minimum when it is a square. 18. Construct equilateral triangles on subsegments AC, CB of segment AB. Prove that if D, D’ are the centers of circles constructed about these triangles, then 6DD’ 2 = AB 2 + AC 2 + CB 2 . 19. If a, b denote the sides of a right triangle about the right angle and p denotes the perpendicular from the right angle on the hypotenuse, then + j_ _ j_ b 2 ~ c 2 ‘ 20. If upon the greater subsegment AB of a segment AC which is divided in extreme and mean ratio, an equilateral triangle AABD is constructed and CD is joined, then CD 2 = 2AB 2 . 21. If a variable line, whose endpoints rest on the circumferences of two given concentric circles, stands opposite a right angle at any fixed point, then the locus of its midpoint is a circle. CHAPTER 3 Circles Axioms and Mathematical Properties from chapters 1 and 2 will generally not be be cited. This will be a rule that we will apply to subsequent chapters, mutatis mutandis. 3.1. Definitions 1. Equal circles are those whose radii are equal. (Note: This is a theorem, and not a definition. If two circles have equal radii, they are evidently congruent figures and therefore equal. Using this method to prove the theorem, [3.26]-[3.29] follow immediately.) 2. A chord of a circle is the segment joining two points on its circumference. 3. A segment, ray, or straight line is said to touch a circle when it intersects the circumference of a circle at one and only one point. The segment, ray, or straight line is called a tangent to the circle, and the point where it touches the circumference is called the point of intersection. FIGURE 3.1.1. [Def. 3.3] CD touches oFEB at B. Or, CD is tangent to oFEB and B is the point of intersection between oFEB and CD. (Note: Modern geometry no longer uses Euclid’s definitions for curves, tan- gents, etc. However, it would be far easier to write a new geometry textbook 123 3.1. DEFINITIONS 124 from first principles rather than attempt to update each of Euclid’s defini- tions and begin agaiiQ However, Euclid’s powerful presentation of complex ideas from simple axioms remains a model for how mathematics should be ap- proached, and students who attempt to master Euclid will find 21st century mathematics more straightforward by doing so.) 4. Circles are said to touch one another when they intersect at one and only one point. There are two types of contact: a) When one circle is external to the other. b) When one circle is internal to, or inside, the other. FIGURE 3.1.2. [Def. 3.4] The circles oBEH and oIEJ touch externally, and the circles oBEH and oBFG touch internally. 5. A segment of a circle is a figure bounded by a chord and one of the arcs into which it divides the circumference. FIGURE 3.1.3. [Def. 3.5] The chord CD of the circle oDEB di- vides the circle into segments DEC and DBC. Segment DEC is bounded by chord CD and arc DEC, and segment DBC is bounded by chord CD and arc DBC. One such attempt is “The Foundations of Geometry” by David Hilbert, http: \/\/www.gutenberg. |org\/ebooks\/17384| 3.1. DEFINITIONS 125 6. Chords are said to be equally distant from the center when the perpen- diculars constructed to them from the center are equal in length. 7. The angle contained by two lines constructed from any point on the circumference of a segment to the endpoints of its chord is called an angle in the segment. 8. The angle of a segment is the angle contained between its chord and the tangent at either endpoint. (Note: A theorem is tacitly assumed in this definition, specifically that the angles which the chord makes with the tangent at its endpoints are equal. We shall prove this later on.) 9. An angle in a segment is said to stand on its conjugate arc. 10. Similar segments of circles are those that contain equal angles. 11. A sector of a circle is formed of two radii and the arc included between them. FIGURE 3.1.4. [Def. 3.11] oDEB with radii AD and AC forms the sectors DACE and DACB. 12. Concentric circles are those which have the same center point. 13. Points which lie on the circumference of a circle are called coney die. 14. A cyclic quadrilateral is one which is inscribed in a circle. 15. A modern definition on an angles In geometry, an angle is the figure formed by two rays called the sides of the angle which share a common endpoint called the vertex of the angle. This measure is the ratio of the length of a circular arc to its radius, where the arc is centered at the vertex and delimited by the sides. The size of a geometric angle is usually characterized by the magnitude of the smallest rotation that maps one of the rays into the other. Angles that have the same size are called congruent angles. ^http : \/\/en.wikipedia. org\/wiki\/Angle 3.1. DEFINITIONS 126 14 l_home_daniel_Documents_Euclid_Project_images-book3_Fig7p FIGURE 3.1.5. The measure of angle 0 is the quotient of s and r. Author: Gustavb, released under the terms of the GNU Free Documentation License, Version 1.2. In order to measure an angle 0, a circular arc centered at the vertex of the angle is constructed, e.g. with a pair of compasses. The length of the arc is then divided by the radius of the arc r, and possibly multiplied by a scaling constant k (which depends on the units of measurement that are chosen): 0 = ks\/r The value of 9 thus defined is independent of the size of the circle: if the length of the radius is changed, then the arc length changes in the same pro- portion, and so the ratio s\/r is unaltered. A number of units are used to represent angles: the radian and the degree are by far the most commonly used. Most units of angular measurement are defined such that one turn (i.e. one full circle) is equal to n units, for some whole number n. In the case of degrees, n = 360. A turn of n units is obtained by setting k = ^ in the formula above. The radian is the angle stands opposite (opposed) by an arc of a circle that has the same length as the circle’s radius (k = 1 when k = ^). One turn is 27r radians, and one radian is 180\/7T degrees, or about 57.2958 degrees. The radian is abbreviated rad, though this symbol is often omitted in mathemati- cal texts, where radians are assumed unless specified otherwise. When radi- ans are used, angles are considered as dimensionless. The radian is used in virtually all mathematical work beyond simple practical geometry, due to the “natural” properties that the trigonometric functions display when their argu- ments are in radians. The radian is the (derived) unit of angular measurement in the SI system. 142_home_daniel_Documents_Euclid_Project_images-book3_Fig7p FIGURE 3.1.6. 0 = s\/r rad = 1 rad. Author: Gustavb, released under the terms of the GNU Free Documentation License, Ver- sion 1.2. The degree, denoted by a small superscript circle (\u00b0), is 1\/360 of a turn, so one turn is 360\u00b0. Fractions of a degree may be written in normal decimal notation (e.g. 3.5\u00b0 for three and a half degrees), but the “minute” and “second” sexagesimal subunits of the “degree-minute-second” system are also in use, especially for geographical coordinates and in astronomy and ballistics. 3.2. PROPOSITIONS FROM BOOK III 127 Although the definition of the measurement of an angle does not support the concept of a negative angle, it is frequently useful to impose a conven- tion that allows positive and negative angular values to represent orientations and\/or rotations in opposite directions relative to some reference. In a two-dimensional Cartesian coordinate system, an angle is typically defined by its two sides, with its vertex at the origin. The initial side is on the positive x-axis, while the other side or terminal side is defined by the measure from the initial side in radians, degrees, or turns. Positive angles represent rotations toward the positive y-axis, and negative angles represent rotations toward the negative y-axis. When Cartesian coordinates are represented by standard position, defined by the x-axis rightward and the y-axis upward, pos- itive rotations are anticlockwise and negative rotations are clockwise. c FIGURE 3.1.7. ZCBA measured as a positive angle, ZEDF measured as a negative angle 3.2. Propositions from Book III PROPOSITION 3.1. THE CENTER OF A CIRCLE I. It is possible to locate the center of a circle. PROOF. We wish to find the center of a given circle (oADB). 3.2. PROPOSITIONS FROM BOOK III 128 Take any two points A, B in the circumference. Join AB and bisect AB at C. Construct CD JL AB and extend CD to intersect the circumference at E. Bisect DE at F . We claim that F is the center of oADB. Suppose instead that point G which does not lie on chord DE is the center of oADB. Join GA, GC, GB. In the triangles AACG, ABCG, we have AC = CB by construction, GA = GB (because they are radii by hypothesis), and the triangles share side CG in common. By [1.8], we have that ZACG = ZBCG. Therefore, each angle is a right angle. But ZACD is right by construction; therefore ZACG = ZACD and ZACG = ZACD + ZDCG, a contradiction. Hence no point can be the center other than a point on chord DE. By the definition of a circle, it follows that F, the midpoint of DE, must be the center ofoADB. \u25a1 Alternatively: PROOF. Because ED bisects AB at a right angle, every point equally dis- tant from the points A, B must lie on ED [1.10, #2]. However, the center is also equally distant from A and B. Hence the center must lie on ED. And since it must be equally distant from E and D, it must be the midpoint of ED. \u25a1 COROLLARY. 1. The line which bisects any chord of a circle perpendicularly passes through the center of the circle. COROLLARY. 2. The locus of the centers of the circles which pass through two fixed points is the line bisecting at right angles the line that connects the two points. 3.2. PROPOSITIONS FROM BOOK III 129 COROLLARY. 3. If A, B, C are three points on the circumference of a circle, the lines which perpendicularly bisect the chords AB, BC intersect in the center of the circle. Proposition 3.2. POINTS ON A LINE INSIDE AND OUTSIDE A CIR- CLE. If any two points are chosen from the circumference of a circle and a line is constructed on these points, then: 1. The points between the endpoints on the circumference form a chord (i.e., they lie inside the circle). 2. The remaining points of the straight line lie outside the circle. PROOF. If any two points (A, B) are taken on the circumference of a circle (oFGH), we claim that: 1. If we construct the line AB, then that segment of the line which lies on and between the points A, B lies within the circle; that is, the line AB contains a segment which is a chord of oFGH. 2. The remaining points of AB lie outside of the circle. Figure 3.2.2. [3.2] We prove each claim separately: 1. Let C be the center of oFGH. Take any point D on the segment AB (as opposed to the line AB) and join CA, CD, and CB. Notice that the angle ZADC is greater than A ABC [1.16]; however, ZABC = ZCAB because ACAB is isosceles [1.5]. Therefore, the angle ZADC > ZCAB; it also follows that ZADC > ZCAD. Hence, AC > CD [1.29], and so CD is less than the radius of oFGH. Consequently, the point D must lie within the circle [Def. 1.23]. Similarly, every other point between A and B lies within oFGH. Finally, since A and B are points in the circumference of oFGH, AB is a chord. 2. Extend the segment AB in both directions and let E be any point on the extension of AB. Wlog, we assume that E lies closer to B than to A. Join 3.2. PROPOSITIONS FROM BOOK III 130 CE. Then the angle ZABC > ZAEC [1.16]; therefore ZCAB > ZAEC. Hence CE > CA, and so the point E lies outside oFGH. \u25a1 COROLLARY. 1. Three collinear points cannot be concyclic. COROLLARY. 2. A straight line, ray, or segment cannot meet a circle at more than two points. COROLLARY. 3. The circumference of a circle is everywhere concave to- wards the center. PROPOSITION 3.3. CHORDS I. Suppose there exist two chords of a circle, one of which passes through the center of the circle. The chord which does not pass through the center is bisected by the chord through the center if and only if the chords are perpendiculars. PROOF. Suppose there exist two chords (AB, CD) of a circle (oACD), one of which passes through the center of the circle (AB). We claim that the other chord (CD) is bisected if and only if AB _L CD. Figure 3.2.3. [3.3] Suppose first that AB bisects CD. We wish to show that AB _L CD. Let O be the center of oACD. Join OC, OD. Then the triangles ACEO, ADEO have CE = ED by hypothesis, OC = OD since each are radii of oACD, and both triangles have EO in common. By [1.8], ZCEO \u2014 ZDEO; since they are also adjacent angles, each angle is a right angle. Hence AB _L CD. Now suppose that AB _L CD. We wish to show that AB bisects CD. Because OC = OD, we have that ZOCD = ZODC by [1.5]. Also ZCEO = ZDEO by hypothesis, since each angle is right. Therefore, the triangles ACEO, ADEO have two angles in one respectively equal to two angles in the other 3.2. PROPOSITIONS FROM BOOK III 131 and the side EO common. By [1.26], CE = ED. Since CD = CE + ED, CD is bisected at E by AB. \u25a1 The second part of the proposition may also be proved in this way: PROOF. By [1.47], we have that OC 2 = OE 2 + EC 2 OD 2 = OE 2 + ED 2 Since we also have that OC 2 = OD 2 , it follows that EC 2 = ED 2 , and so EC = ED. \u25a1 COROLLARY. 1. The line which bisects perpendicularly one of two parallel chords of a circle bisects the other perpendicularly. COROLLARY. 2. The locus of the midpoints of a system of parallel chords of a circle is the diameter of the circle perpendicular to them all. COROLLARY. 3. If a line intersects two concentric circles, its intercepts be- tween the circles are equal in length. COROLLARY. 4. The line joining the centers of two intersecting circles bi- sects their common chord perpendicularly. Observation: [3.1], [3.3], and [3.3, Cor. 1] are related such that if any one of them is proved directly, then the other two follow by the Rule of Symmetry. Exercises. 1. If a chord of a circle stands opposite a right angle at a given point, the locus of its midpoint is a circle. 2. Prove [3.3, Cor. 1]. 3. Prove [3.3, Cor. 4]. PROPOSITION 3.4. CHORDS II. If two chords, one of which is not a diam- eter, intersect one another in a circle, they do not bisect each another. 3.2. PROPOSITIONS FROM BOOK III 132 PROOF. In a circle oACB, construct two chords (AB, CD) which are not both diameters and intersect each other at a point (E). We claim that AB and CD do not bisect each other. Figure 3.2.4. [3.4] Let O be the center of oACB. Since AB, CD are not both diameters, we may join OE. Suppose that AE = EB and CE = ED. Since OE, which intersects the center of the circle, bisects AB, which does not intersect the center of the circle, we must have that OE _L AB. Similarly, OE _L CD. Hence, ZAEO = ZCEO where ZCEO = ACE A + ZAEO, a contradiction. Therefore, AB and CD do not bisect each other. \u25a1 COROLLARY. 1. If two chords of a circle bisect each other, they are both diameters. (This is the contrapositive statement of [3.4].) PROPOSITION 3.5. NON-CONCENTRIC CIRCLES I. If two circles inter- sect one another at two points, they are not concentric. (See [Def. 2.13].) PROOF. If two circles (oABC, oABD) intersect at two points (A and B), we claim that the circles are not concentric. 3.2. PROPOSITIONS FROM BOOK III 133 Suppose that oABC, oABD share a common center, O. Join OA and con- struct a segment OD (where B ^ D) which cuts the circles at C and D, respec- tively. Because O is the center of the circle oABC, OA = OC. Because O is the center of the circle oABD, OA = OD. Hence, OC = OD and OD = OC + CD, a contradiction. Therefore, oABC, oABD are not concentric. \u25a1 Exercises. 1. If two non-concentric circles intersect at one point, they must intersect at another point. For let O, O’ be the centers of these circles and A be their point of intersection. From A, let AC be the perpendicular on the segment 00′ . Extend AC to B, making BC = CA. It follows that B is another point of intersection. 2. Two circles cannot have three points in common without coinciding. PROPOSITION 3.6. NON-CONCENTRIC CIRCLES II. If one circle inter- sects another circle internally at one and only one point, then the circles are not concentric. PROOF. If a circle (oABC) touches another circle (oADE) internally at one and only one point (A), then the circles are not concentric. 3.2. PROPOSITIONS FROM BOOK III 134 Figure 3.2.6. [3.6] To prove this, suppose that the circles are concentric and let O be the cen- ter of each. Join OA, and construct any other segment OD, cutting the circles at the points B, D respectively. Because O is the center of each circle by hy- pothesis, OA = OB and OA = OD; therefore, OB = OD and OB + BD = OD, a contradiction. Hence, the circles are not concentric. \u25a1 PROPOSITION 3.7. UNIQUENESS OF SEGMENT LENGTHS FROM A POINT ON THE DIAMETER OTHER THAN THE CENTER. If on the diame- ter of a circle a point is taken (other than the center of the circle) and from that point segments are constructed to the circumference, the longest segment will contain the center of the circle and the shortest segment will form a diameter with the longest segment. As for the remaining segments, those with endpoints on the circumference nearer to the endpoint on the circumference of the longest segment will be longer than segments with endpoints on the circumference far- ther from the endpoint of the longest segment. Also, only two equal straight-line segments may be constructed from that point to the circumference, one on each side of the least segment constructed from the given point to circumference. PROOF. If from any point (P) on a diameter of a circle (other than the center, O) we construct segments {PA, PB, PC, etc.) to the circumference, one of which passes through the center (PA), we claim that: 1. The longest is the segment which passes through the center (PA). 2. The extension of this segment in the opposite direction (PE) is the short- est segment. 3. Of the others, the segment which is nearest to the segment which passes through the center (PA) is greater than every segment which is more remote (i.e., PA> PB > PC > PD). 3.2. PROPOSITIONS FROM BOOK III 135 4. Any two segments making equal angles with the diameter and on oppo- site sides of the diameter are equal in length (i.e., PD = PF). 5. More than two equal segments cannot be constructed from the given point (P) to the circumference. Figure 3.2.7. [3.7] oEAG We prove each claim separately: 1. The longest is the segment which passes through the center (PA). Let O be the center of oEAG. Join OB. Clearly, OA = OB. From this we obtain PA = OA + OP = OB + OP. Since OB + OP> PB [1.20], it follows that PA > PB. 2. The extension of this segment in the opposite direction (PE) is the short- est segment. Join OD. By [1.20], OP + PD > OD. Since, OD = OE, it follows that OP + PD > OE. Subtracting OP from each side of the inequality, we have that PD > PE. 3. Of the others, the segment which is nearest to the segment which passes through the center (PA) is greater than every segment which is more remote (i.e., PA> PB > PC > PD). Join OC. The two triangles APOB, APOC have the sides OB = OC with OP in common. But the angle ZPOB > ZPOC since ZPOB = ZPOC + ZBOC. By [1.24], we have that PB > PC. Similarly, PC > PD. 4. Any two segments making equal angles with the diameter and on oppo- site sides of the diameter are equal in length (i.e., PD = PF). 3.2. PROPOSITIONS FROM BOOK III 136 At O, construct ZPOF = ZPOD and join PF. Consider the triangles APOD, APOF: each shares side OP, OD = OF, and ZPOD = ZPOF by con- struction. By [1.4], APOD ^ APOF, and so ZOPF = ZOPD and PD = PF. The proof follows. 5. More than two equal segments cannot be constructed from the given point (P) to the circumference. We claim that a third segment cannot be constructed from P equal to PD = PF. Suppose this were possible and let PG = PD. Then PG = PF; that is, the segment which is nearest to the segment through the center is equal to the one which is more remote; this contradicts point 3, above. Hence, three equal segments cannot is constructed from P to the circumference. \u25a1 COROLLARY. 1. If two equal segments PD, PF are constructed from a point P to the circumference of a circle, the diameter through P bisects the angle ZDPF formed by these segments. COROLLARY. 2. If P is the common center of circles whose radii are PA, PB, PG, PD, etc., then: (a) The circle whose radius is the maximum segment (PA) lies outside the circle oADE and touches it at A [Def. 3.4]. (b) The circle whose radius is the minimum segment (PE) lies inside the circle oADE and touches it at E. (c) A circle having any of the remaining radii (such as PD) cuts oADE at two points ( D, F). Observation: [3.7] is a good illustration of the following important defini- tion: if a geometrical magnitude varies its position continuously according to any well-defined relationship, and if it retains the same value throughout, it is said to be a constant (such as the radius of a fixed circle). But if a magnitude increases for some time and then begins to decrease, it is said to be a maxi- mum at the end of the increase. Therefore in the previous figure, PA, which we suppose to revolve around P and meet the circle, is a maximum. Again, if it decreases for some time, and then begins to increase, it is a minimum at the beginning of the increase. Thus PE, which we suppose as before to re- volve around P and meet the circle, is a minimum. [3.8] will provide other illustrations of this concept. 3.2. PROPOSITIONS FROM BOOK III 137 Proposition 3.8. SEGMENT LENGTHS FROM A POINT OUTSIDE THE CIRCLE AND THEIR UNIQUENESS. Suppose a point is chosen outside of a circle and from that point segments are constructed such that they intersect the circumference of the circle at two points, one on the “outer” or convex side of the circumference and one on the “inner” or concave side of the circumference. Let one segment be constructed which intersects the center of the circle and the others all within the same semicircle but not through the center of the circle. Then: 1. The maximum segment passes through the center. 2. Of the others, those nearer to the segment through the center are greater in length than those which are more remote. 3. If segments are constructed to the convex circumference, the minimum segment is that which passes through the center when extended. 4. Of the other segments, that which is nearer to the minimum is less than one more remote. 5. From the given point outside of the circle, there can be constructed two equal segments to the concave or the convex circumference, both of which make equal angles with the line passing through the center. 6. Three or more equal segments cannot be constructed from the given point outside the circle to either circumference. PROOF. Construct oADK and point P outside of the circle. We prove each claim separately: \u00a5 Figure 3.2.8. [3.8] oADK 1. The maximum segment passes through the center. 3.2. PROPOSITIONS FROM BOOK III 138 Let O be the center of oADK and join OB. Clearly, OA = OB. From this we obtain AP = OA + OP = OB + OP. But the sum OB + OP > BP [1.20]. Therefore, > BP. 2. Construct the remaining segments from the figure. Those nearer to the segment through the center are greater in length than those which are more remote. Join OC, OD. The two triangles ABOP, ACOP have the side OB equal in length to OC with OP in common, and the angle ZBOP > ACOP. Therefore, BP > CP [1.24]. Similarly, CP > DP, etc. 3. If segments are constructed to the convex circumference, the minimum segment is that which passes through the center when extended. Join OF. In AOFP, the sum OF + FP > OP = OE + EP [1.20]. Recall that OF = OE. Subtracting OE and OF from each side of the inequality, we have that FP > EP. 4. Of the other segments, that which is nearer to the minimum is less than one more remote. Join OG, OH. The two triangles AGOP, AFOP have two sides GO, OP in one respectively equal to two sides FO, OP in the other, and the angle ZGOP > ZFOP By [1.24], GP > FP. Similarly, HP > GP. 5. From the given point outside of the circle, there can be constructed two equal segments to the concave or the convex circumference, both of which make equal angles with the line passing through the center. Construct the angles ZPOI = ZPOF [1.23] and join IP. The triangles AIOP, AFOP have two sides 10, OP in one respectively equal to two sides FO, OP in the other, and ZIOP = ZFOP by construction. By [1.4], IP = FP. 6. Three or more equal segments cannot be constructed from the given point outside the circle to either circumference. A third segment cannot is constructed from P equal to either of the seg- ments IP, FP. If this were possible, let PK = PF. Then PK = PI, which contradicts part 4, above. \u25a1 COROLLARY. 1. If PI is extended to meet the circle at L, then PL = PB. 3.2. PROPOSITIONS FROM BOOK III 139 COROLLARY. 2. If two equal segments are constructed from P to either the convex or concave circumference, the diameter through P bisects the angle between them, and the segments intercepted by the circle are equal in length. COROLLARY. 3. If P is the common center of circles whose radii are seg- ments constructed from P to the circumference of oADK, then: a) The circle whose radius is the minimum segment (PE) has external con- tact with oADK [Def 3.4]. b) The circle whose radius is the maximum segment (PA) has internal con- tact with oADK. c) A circle having any of the remaining segments (PF) as radius intersects oADK at two points (F, I). PROPOSITION 3.9. THE CENTER OF A CIRCLE II. A point within a circle from which three or more equal segments can be constructed to the circumfer- ence is the center of that circle. PROOF. Let Dbea point within oABC and from D construct equal seg- ments DA, DB, DC which intersect the circumference. We claim that D is the center of oABC. Figure 3.2.9. [3.9] Join AB, BC and bisect them at points E and F, respectively. Join ED and FD and extend these segments to the points G, K, H, and L on the circumfer- ence [1.10]. Since AE = EB and ED is a common side, the two sides AE, ED of AAED equal the sides BE, ED of ABED; we also have that the base DA 3.2. PROPOSITIONS FROM BOOK III 140 equals the base DB since each are radii of oABC. By [1.8], AAED ^ ABED, and so ZAED = ABED. It follows that AAED, ABED are each right angles. We have that GK cuts AB perpendicularly into two equal parts. By [3.1, Cor. 1], the center of oABC is a point on GK. Similarly, the center of oABC is a point on HL. Since GK and HL have no other point of intersection except for D, it follows that D is the center of oABC. \u25a1 Alternatively: PROOF. Since AD = LD, the segment bisecting the angle ZADL passes through the center [3.7, Cor. 1]. Similarly, the segment bisecting the angle ZBDA passes through the center. Hence, the point of intersection of these bisectors, D, is the center. \u25a1 PROPOSITION 3.10. THE UNIQUENESS OF CIRCLES. If two circles have more than two points of their circumferences in common, they coincide. PROOF. If two circles (oABC, oDAB) have more than two points of the circumference in common, they coincide. Figure 3.2.10. [3.10] To prove this, suppose that oABC , oDAB share three points in common (A, B, C) without coinciding. Locate P, the center of oABC. Join PA, PB, PC. Since P is the center of oABC, we have that PA = PB = PC. Again, since oDAB is a circle and P a point from which three equal lines PA, PB, PC can be constructed to its circumference, P must be the center of oDAB [3.9]. Hence oABC and oDAB are concentric, a contradiction. \u25a1 COROLLARY. 1. Two circles which do not coincide do not have more than two points common. 3.2. PROPOSITIONS FROM BOOK III 141 Note: Similarly to [3.10, Cor. 1], two lines which do not coincide cannot have more than one point common. Proposition 3.11. SEGMENTS CONTAINING CENTERS OF CIRCLES. If one circle touches another circle internally at one point, then the segment joining the centers of the two circles and terminating on the circumference must have that point of intersection as an endpoint. PROOF. If a circle (oCPD) touches another circle (oAPB) internally at a point (P), we claim that the segment joining the centers of these circles has P as an endpoint. Figure 3.2.11. [3.11] Let O be the center of oAPB and join OP. We claim that the center of oCPD is a point on the segment OP. Otherwise, let the center of oCPD be other than on OP; wlog, choose point E. Join OE, EP, and extend OE through E to meet oCPD at C and oAPB at A. Since E is a point on the diameter of oAPB between O and A, EA < EP [3.7]. But EP = EC by hypothesis since they are radii of oCPD. Hence EA AC + DB. Therefore, AB > AP + PB, and one side of AAPB is greater than the sum of the other two, a contradiction [1.20]. Hence the center of oPDE must be on the segment PE. Let it be G, and the proof follows. \u25a1 Alternatively: PROOF. Suppose that the center of oPDE lies on the segment BP. Since BP is a segment constructed from a point outside of the circle oPCF to its circumference which does not pass through the center when it is extended, 3.2. PROPOSITIONS FROM BOOK III 143 the circle whose center is B with radius BP must cut the circle oPCF at P [1.8, Cor. 3]. However, it touches oPCF at P by hypothesis, a contradiction. Since BP was chosen arbitrarily, the center of oPDE must lie on the segment PE. \u25a1 Observation: [3.11] and [3.12] may both be included in one theorem: “If two circles touch each other at any point, the centers and that point are collinear.” This is a limiting case of the theorem given in [3.3, Cor. 4]: “The line joining the centers of two intersecting circles bisects the common chord perpendicularly.” FIGURE 3.2.13. [3.12], Observation. Suppose the circles with centers O and O’ have two points of intersection, A and B. Sup- pose further that A remains fixed while the second circle moves so that the point B ultimately coincides with A. Since the seg- ment 00′ always bisects AB, we see that 00′ intersects A. In consequence of this motion, the common chord CD becomes the tangent to each circle at A. COROLLARY. 1. If two circles touch each other, their point of intersection is the union of two points of intersection. When counting the number of points at which two circles intersect, we may for purposes of calculation consider this point of intersection as two points. See Cor. 2 for details. 3.2. PROPOSITIONS FROM BOOK III COROLLARY. 2. If two circles touch each other at any point, have any other common point. 144 they cannot FIGURE 3.2.14. [3.2, Cor. 2] For, since two circles cannot have more than two points common [3.10] and their point of intersection is equivalent to two points for purposes of calculation, circles that touch cannot have any other point common. The following is a formal proof of this Corollary: Let O, O’ be the centers of the two circles where A is the point of intersection, and let O’ lie between O and A. Take any other point B in the circumference of O, and join O’B. By [3.7], O’B > O’ A. Therefore, B is outside the circumference of the inner circle. Hence, B cannot be common to both circles. Since point B was chosen arbitrarily, the circles cannot have any other common point except for A. PROPOSITION 3.13. INTERSECTING CIRCLES II. Two circles cannot touch each other at two points either internally or externally. PROOF. We divide the proof into its internal and external cases: 3.2. PROPOSITIONS FROM BOOK III 145 Figure 3.2.15. [3.13] 1. Suppose two distinct circles oACB and oADB touch internally at two points A and B. Since the two circles touch at A, the segment joining their centers passes through A [3.11]. Similarly, the segment joining their centers passes through B. Hence, the centers of these circles and the points A, B are on one segment, and so AB is a diameter of each circle. Hence, if AB is bisected at E, E must be the center of each circle, i.e., the circles are concentric, a contradiction [1.5]. 2. If two circles oACB and oIJK touch externally at points \/ and J, then by [3.12] the segment joining the centers of oACB and oIJK passes through the centers E and H as well as points \/ and J, a contradiction. \u25a1 An alternative proof to part 1: PROOF. Construct a line bisecting AB perpendicularly. By [3.1, Cor. 1], this line passes through the center of each circle, and by [3.11], [3.12] must pass through each point of intersection, a contradiction. Hence, two circles cannot touch each other at two points. \u25a1 This proposition is an immediate inference from [3.12, Cor. 1] that a point of intersection counts for two intersections, for then two contacts would be equivalent to four intersections; but there cannot be more than two intersec- tions [3.10]. It also follows from [3.12, Cor. 2] that if two circles touch each other at point A, they cannot have any other point common. Hence, they can- not touch again at B. 3.2. PROPOSITIONS FROM BOOK III 146 Exercises. 1. If a circle with a non-fixed center touches two fixed circles externally, the difference between the distances of its center from the centers of the fixed circles is equal to the difference or the sum of their radii, according to whether the contacts are of the same or of opposite type [Def. 3.4]. 2. If a circle with a non-fixed center is touched by one of two fixed circles internally and touches the other fixed circle either externally or internally, the sum of the distances from its center to the centers of the fixed circles is equal to the sum or the difference of their radii, according to whether the contact with the second circle is internal or external. 3. Suppose two circles touch externally. If through the point of intersec- tion any secant is constructed cutting the circles again at two points, the radii constructed to these points []e parallel. 4. Suppose two circles touch externally. If two diameters in these circles are parallel, the line from the point of intersection to the endpoint of one di- ameter passes through the endpoint of the other. 5. Rewrite the results of #3 and #4. PROPOSITION 3.14. EQUALITY OF CHORD LENGTHS. Chords in a cir- cle are equal in length if and only if they are equally distant from the center. PROOF. In circles with equal radii, we claim that: 1. chords of equal length (AB, CD) are equally distant from the center (O). 2. chords which are equally distant from the center are also equal in length. Figure 3.2.16. [3.14] Let O be the center of oACD, and construct the chords AB, CD. We prove each claim separately: 1. Let AB = CD and construct perpendicular segments OE, OF. We wish to prove that OE = OF. 3.2. PROPOSITIONS FROM BOOK III 147 Join AO, CO. Because AB is a chord in a circle and OE is a segment constructed from the center where OE _L AB, OE bisects AB [3.3]. It follows that AE = EB. Similarly, CF = FD. But AB = CD by hypothesis, and so AE = CF. Because ZOEF is a right angle, AO 2 = AE 2 + EO 2 . Similarly, CO 2 = CF 2 + FO 2 . Since AO 2 = CO 2 , we have that AF 2 + EO 2 = CF 2 + FO 2 where AF 2 = CF 2 . Hence EO 2 = FO 2 , and so EO = FO. Hence AB, CD are equally distant from O [Def. 3.6]. 2. Now let EO = FO. We wish to prove that AB = CD. As before, we have AE 2 + FO 2 = CF 2 + FO 2 where FO 2 = FO 2 by hypoth- esis. Hence AE 2 = CF 2 , and so AE = CF. But AB = 2 \u25a0 AE and CD = 2 \u25a0 CF, from which it follows that AB = CD. \u25a1 Exercise. 1. If a chord of given length slides around a fixed circle, then: a) the locus of its midpoint is a circle; b) the locus of any point fixed on the chord is a circle. PROPOSITION 3.15. INEQUALITY OF CHORD LENGTHS. The diameter is the longest chord in a circle, and a chord is nearer to the center of a circle than another chord if and only if it is the longer of the two chords. PROOF. Construct oACB with center O, diameter AB, and chords CD, EF. We claim that: 1. The diameter (AB) is the longest chord in a circle; 2. A chord which is nearer to the center (CD) is longer than a chord which is more distant (EF); 3. Longer chords are nearer to the center than shorter chords. Figure 3.2.17. [3.15] We prove each claim separately: 3.2. PROPOSITIONS FROM BOOK III 148 1. The diameter (AB) is the longest chord in a circle. Join OC, OD, OE and construct the perpendiculars OG, OH. Because O is the center of oAEF, OA = OC and OB = OD. Hence AB = OA + OB = OC + OD. But OC + CD > CD by [1.20]. Therefore, AB > CD. 2. A chord which is nearer to the center (CD) is longer than a chord which is more distant (EF). Suppose that CD is nearer to O than EF. It follows that OG < OH [3.14]. Since AOCC and AOHE are right triangles, we have that OC 2 = OG 2 + CC 2 and OE 2 = OH 2 + #\u00a3 2 . Since OC = 0\u00a3, OC 2 + GC 2 = OH 2 + #\u00a3 2 . But OG 2 #\u00a3 2 . It follows that GC > HE. Since CD = 2 \u2022 GC and EF = 2- HE, it follows that CD > EF. 3. Longer chords are nearer to the center than shorter chords. Suppose that CD > EF. We wish to prove that OG HE 2 . Therefore OG 2 < OH 2 , and so OG CB 2 , and so CI > CB. This confirms that \/ lies outside of the circle oDAB [3.2]. Similarly every other point on BI except B lies outside of the oDAB. Hence, since BI meets the circle oDAB at B and does not cut it, BI must touch oDAB at B. 2. Any other line or segment (BH) through the same point (B) cuts the circle. We wish to prove that BH, which is not perpendicular to AB, cuts the circle. Construct CG _L HB. It follows that BC 2 = CG 2 + GB 2 . Therefore BC 2 > CG 2 , and so BC > CG. Hence the point G must be within the circle [3.2], and consequently if the segment BG is extended it must meet oDAB at H and therefore cut it. \u25a1 Exercises. 1. If two circles are concentric, all chords of the greater circle which touch the lesser circle are equal in length. 2. Construct a parallel to a given line which touches a given circle. 3. Construct a perpendicular to a given line which touches a given circle. 4. Construct a circle having its center at a given point a) and touches a given line; b) and touches a given circle. 3.2. PROPOSITIONS FROM BOOK III 150 How many solutions exist in this case? 5. Construct a circle of given radius that touches two given lines. How many solutions exist? 6. Find the locus of the centers of a system of circles touching two given lines. 7. Construct a circle of given radius that touches a given circle and a given line or that touches two given circles. PROPOSITION 3.17. TANGENTS ON CIRCLES I. It is possible to construct a tangent of a given circle from a given point outside of the circle. PROOF. We wish to construct a tangent to a given circle (oBCD) from a given point (P) outside of the circle. Figure 3.2.19. [3.17] (a), ((3) Let O be the center of oBCD (Fig. 3.2.19). Join OP, cutting the circumfer- ence at C. With O as center and OP as radius, construct the circle oEAP. Also construct CA _L OP. Join OA, intersecting oBCD at B, and join BP. We claim that BP is the required tangent to oBCD.O Since O is the center of oCDB and oEAP, we have that OA = OP and OC = OB. Hence AAOC, APOB have the sides OA, OC in one triangle re- spectively equal to the sides OP, OB in the other with the contained angle common to both. By [1.4], AAOC ^ APOB, and so \/OCA = ZOBP. But Z.OCA is a right angle by construction. Therefore ZOBP is a right angle, and so by [3.16], PB touches the circle oBCD at B. Hence, PB is a tangent of oBCD at point B. \u25a1 3.2. PROPOSITIONS FROM BOOK III 151 COROLLARY. 1. If AC in Fig. 3.2.26(j3) is extended to E, OE is joined, the circle oBCD is cut at D, and the segment DP is constructed, then DP is another tangent ofoBCD (at point P). Exercises. 1. The two tangents PB, PD (in Fig. 3.2.26(\/3)) are equal in length to one another because the square of each is equal to the square of OP minus the square of the radius. 2. If a quadrilateral is circumscribed to a circle, the sum of one pair of opposite sides is equal to the sum of the other pair. 3. If a parallelogram is circumscribed to a circle, it must be a lozenge, and its diagonals intersect at the center. 4. In Fig. 3.2.26(\/3), if BD is joined and OP is intersected at F, then OP _L BD. 5. The locus of the intersection of two equal tangents to two circles is a segment (called the radical axis of the two circles). 6. Find a point such that tangents from it to three given circles is equal. (This point is called the radical center of the three circles.) 7. The rectangle OF.OP is equal in area to the square of the radius of oBCD. Definition: Suppose we have two points F and P such that when the area of the rectangle OF.OP (where O is the center of a given circle) is equal to the area of the square of the radius of that circle, then F and P are called inverse points with respect to the circle. 9. The intercept made on a variable tangent by two fixed tangents stands opposite a constant angle at the center. 10. Construct a common tangent to two circles. Demonstrate how to con- struct a segment cutting two circles so that the intercepted chords are of given lengths. PROPOSITION 3.18. TANGENTS ON CIRCLES II. If a line touches a cir- cle, the segment from the center of the circle to the point of intersection with the line is perpendicular to the line. PROOF. If the line CD touches oABC, we claim that the segment OC from the center (O) to the point of intersection (C) is perpendicular to CD. 3.2. PROPOSITIONS FROM BOOK III 152 A Figure 3.2.20. [3.18] Otherwise, suppose that another segment OG is constructed from the cen- ter such that OG _L CD. Let OG cut the circle at F. Because the angle ZOGC is right by hypothesis, the angle ZOCG must be acute [1.17]. By [1.19], OC > OG. But OC = OF, and therefore we have that OF > OG and OG = OF + FG, a contradiction. Hence OC _L CD. \u25a1 Alternatively: PROOF. Since the perpendicular must be the shortest segment from O to CD and OC is evidently the shortest line, it follows that OC _L CD. \u25a1 PROPOSITION 3.19. TANGENTS ON CIRCLES III. If a line is a tangent to a circle, then the perpendicular constructed from its point of intersection passes through the center of the circle. PROOF. If a line (AB) is tangent to a circle (oCDA), we claim that the per- pendicular (AC) constructed from its point of intersection (A) passes through the center of oCDA. Figure 3.2.21. [3.19] 3.2. PROPOSITIONS FROM BOOK III 153 Suppose it were otherwise; let O be the center of oCDA and join AO. Be- cause the line AB touches oCDA and OA is constructed from the center to the point of intersection, OA _L AB [3.18]. Therefore ZOAB and ZCAB are right angles. It follows that ZOAB = ZCAB and ZCAB = ZOAB + ZOAC, a contradiction. Hence, the center must lie on the segment AC. \u25a1 COROLLARY. 1. If a number of circles touch the same line at the same point, the locus of their centers is the perpendicular to the line at the point. COROLLARY. 2. Suppose we have a circle and any two of the following properties: a) a tangent to a circumference; b) a segment, ray, or straight line constructed from the center of the circle to the point of intersection; c) right angles at the point of intersection. Then by [3.16], [3.18], [3.19], and the Rule of Symmetry, the remaining property follows. If we have (a) and (c), then it may be necessary to extend a given segment or a ray to the center of the circle. These are limiting cases of [3.1, Cor. l]and [3.3]. Proposition 3.20. ANGLES AT THE CENTER OF A CIRCLE AND ON THE CIRCUMFERENCE. The angle at the center of a circle is double the angle at the circumference when each stands on the same arc of the circumference. PROOF. Construct oABC with center O and radius OB as in Fig. 3.2.22 (a). A E A Figure 3.2.22. [3.20], (a), (7) 3.2. PROPOSITIONS FROM BOOK III 154 Construct ZAOB at its center and ZACB at the circumference such that each angle stands on the arc AB. We claim that ZAOB = 2 \u2022 ZACB. Join CO and extend CO through to the circumference at E. Since OC = OA, we have that ZOCA = ZOAC. By [1.5], ZOCA + ZOAC = 2 \u2022 ZOCA Also, ZAO\u00a3 = ZOCA + ZOAC [1.32]. It follows that ZAO\u00a3 = 2 \u2022 ZOCA. Similarly, Z\u00a30\u00a3 = 2 \u2022 ZOC\u00a3, and so we have that ZAOB = ZAOE + Z\u00a30\u00a3 = 2 \u2022 ZOCA + 2 \u2022 ZOCB = 2 \u2022 ZACB Now construct oABC with center O and radius 05 with ZACB as in Fig. 3.2.22 (7). Join CO through to the circumference at E. Similarly to the above, we can show that ZEOB = 2 \u2022 ZOCB and ZEOA = 2 \u2022 ZECA. It follows that ZEOB – ZEOA = ZAOB = 2 \u2022 ZOCB – 2 \u2022 ZECA = 2 \u2022 (ZOCB – ZECA) = 2 \u2022 ZACB Therefore, the angle at the center of a circle is double the angle at the circumference when the angles stand on the same arc. \u25a1 Alternatively: PROOF. Construct oABC as in Fig. 3.2.2(a), oACB as in Fig. 3.2.2(\/3), and oECB in Fig. 3.22( 7 ). Join CO and extend it to E. Because OA = OC, it follows that ZACO = ZOAC. Since ZAOE = ZOAC + ZACO [1.32], we have that ZAOE = 2- ZACO. Similarly, ZEOB = 2 \u2022 ZOCB. Hence (by adding in the cases of Fig 3.2.22 (a), (13), and subtracting in (7)), we have that ZAOB = 2 \u2022 ZACB. \u25a1 COROLLARY. 1. If AOB is a line, then ZACB is a right angle; specifically, the angle in a semicircle is a right angle (compare with [3.31]). PROPOSITION 3.21. ANGLES ON CHORDS. In a circle, angles standing on the same arc are equal in measure to one another. PROOF. We claim that the angles (ZACB, ZADB) standing within oABC and on the same arc (AB) are equal in measure. 3.2. PROPOSITIONS FROM BOOK III 155 Figure 3.2.23. [3.21], (a), (P) We first consider the cases presented in Fig. 3.2.23(a), ((3). Let O be the center of o ABC, and join OA, OB. By [3.20], ZAOB = 2 \u2022 ZACB and ZAOB = 2 \u2022 ZADB. It follows that ZACB = ZADB. We now consider the case presented in Fig. 3.2.24(7): Figure 3.2.24. [3.21] ( 7 ) Let O remain the center of oABC. Join CO and extend the segment to intersect the circumference of oCAB at E. Join DE. Since O is the center, the arc ACE is greater than a semicircle; similarly to our first case, we obtain that ZACE = ZADE and ZECB = ZEDB. Hence, we have that ZACB = ZACE + ZECB = ZADE + ZEDB = ZADB \u25a1 COROLLARY. 1. If two triangles \/ACB, AADB stand on the same base AB and have equal vertical angles on the same side of it, the four points A, C, D, B are concyclic. 3.2. PROPOSITIONS FROM BOOK III 156 COROLLARY. 2. If A, B are two fixed points and if C varies its position in such a way that the angle ZACB retains the same value throughout, the locus of C is a circle. (Or: given the base of a triangle and the vertical angle, the locus of the vertex is a circle). Exercises. 1. Given the base of a triangle and the vertical angle, find the locus (a) of the intersection of its perpendiculars; (b) of the intersection of the internal bisectors of its base angles; (c) of the intersection of the external bisectors of the base angles; (d) of the intersection of the external bisector of one base angle and the internal bisector of the other. 2. If the sum of the squares of two segments is given, their sum is a maxi- mum when the segments are equal in length. 3. Of all triangles having the same base and vertical angle, the sum of the sides of an isosceles triangle is a maximum. 4. Of all triangles inscribed in a circle, the equilateral triangle has the maximum perimeter. 5. Of all concyclic figures having a given number of sides, the area is a maximum when the sides are equal. Proposition 3.22. QUADRILATERALS INSCRIBED INSIDE CIRCLES. The sum of the opposite angles of a quadrilateral inscribed in a circle equals two right angles. PROOF. We claim that the sum of the opposite angles of a quadrilateral (ABCD) inscribed in a circle (oCBA) equals two right angles. Figure 3.2.25. [3.22] 3.2. PROPOSITIONS FROM BOOK III 157 Join AC, BD. Since ZABD and ZACD stand on the same arc AD, we have that ZABD = ZACD [3.21]. Similarly, ZDBC = ZD AC because they stand on the arc DC. Hence, ZABC = ZACD + ZD AC. From this, we obtain ZABC + ZCDA = ZACD + ZD AC + ZCDA where the right-hand side of the equality is the sum of the three angles of AACD. Since this sum equals two right angles [1.32], we have that ZABC + ZCDA equals two right angles. We obtain an analogous result for ZDAB + ZBCD. \u25a1 Alternatively: PROOF. Let O be the center of oCBA. FIGURE 3.2.26. [5.22], alternative proofs Join OA, OC. Define ZAOC as less than two right angles and ZCOA as more than two right angles. Also, notice that ZAOC + ZCOA =four right angles. Notice that ZAOC = 2 \u2022 ZCDA and ZCOA = 2 \u2022 ZABC by [3.20]. Hence ZAOC + ZCOA = 2-(ZCDA + ZABC). Since, ZAOC + ZCOA equals four right angles, ZCDA + ZA\u00a3?C equals two right angles. \u25a1 COROLLARY. 1. If the sum of two opposite angles of a quadrilateral are equal to two right angles, then a circle may be inscribed about the quadrilateral. COROLLARY. 2. If a parallelogram is inscribed in a circle, then it is a rectangle. 3.2. PROPOSITIONS FROM BOOK III 158 Exercises. 1. If the opposite angles of a quadrilateral are supplemental, it is cyclic. 2. A segment which makes equal angles with one pair of opposite sides of a cyclic quadrilateral makes equal angles with the remaining pair and with the diagonals. 3. If two opposite sides of a cyclic quadrilateral are extended to meet and a perpendicular falls on the bisector of the angle between them from the point of intersection of the diagonals, this perpendicular will bisect the angle between the diagonals. 4. If two pairs of opposite sides of a cyclic hexagon are respectively parallel to each other, the remaining pair of sides are also parallel. 5. If two circles intersect at the points A, B, and any two segments ACD, BFE are constructed through A and B, cutting one of the circles in the points C, E and the other in the points D, F, then CE \\ DF. 6. If equilateral triangles are constructed on the sides of any triangle, the segments joining the vertices of the original triangle to the opposite vertices of the equilateral triangles are concurrent. 7. In the same case as #7, prove that the centers of the circles constructed about the equilateral triangles form another equilateral triangle. 8. If a quadrilateral is constructed about a circle, the angles at the center standing opposite the opposite sides are supplemental. 9. If a tangent which varies in position meets two parallel tangents, it stands opposite a right angle at the center. 10. If a hexagon is circumscribed about a circle, the sum of the angles standing opposite the center from any three alternate sides is equal to two right angles. PROPOSITION 3.23. UNIQUENESS OF ARCS. It is impossible to construct two similar and unequal arcs on the same side of the same chord. PROOF. Two similar arcs (ACB, ADB) which do not coincide cannot be constructed on the same of the same chord (AB). Figure 3.2.27. [3.23] 3.2. PROPOSITIONS FROM BOOK III 159 Suppose that we have arcs ACB ~ ADB which are constructed on the same side of segment AB. Take any point D in the inner arc (ADB). Join AD, and extend it to meet the outer arc at C. Also join BC, BD. Then since the arcs are similar, Z ADB = ZACB [Def. 3.10], which contradicts [1.16.]. Hence, the proof. \u25a1 PROPOSITION 3.24. SIMILAR ARCS. Similar arcs standing on equal chords are equal in length. PROOF. We claim that similar arc of circles (AEB, CFD) on equal chords (AB, CD) are equal in length. e = Figure 3.2.28. [3.24] Since AB = CD, if AB is applied to CD such that the point A coincides with C and the chord AB with CD, the point B coincides with D. Because AEB ~ CFD, they must coincide [3.23]. Hence, the proof. \u25a1 COROLLARY. 1. Since the chords are equal in length, they are congruent; therefore the arcs, being similar, are also congruent. Proposition 3.25. CONSTRUCTION OF A CIRCLE FROM AN ARC. Given an arc of a circle, it is possible to construct the circle to which the arc belongs. PROOF. Given an arc (ABC) of a circle, we wish to construct oABC. Figure 3.2.29. [3.25] 3.2. PROPOSITIONS FROM BOOK III 160 Take any three points A, B, C of the arc. Join AB, BC. Bisect AB at D and BC at E. Construct DF, EF at right angles to AB, BC. We claim that F, their point of intersection, is the center of the required circle. Because DF bisects the chord AB and is perpendicular to it, the center of the circle of which ABC is an arc must lie on DF [3.1, Cor. 1]. Similarly, the center of the circle of which ABC is an arc must lie on EF. Hence the point F is the center of oABC (that is, the circle constructed with F as center and FA as radius). \u25a1 Propositions [3.26]-[3.29] are related in the following sense: In [3.26], given equal angles, we must prove that we have equal arcs. In [3.27], given equal arcs, we must prove that we have equal angles. Hence, [3.27] is the converse of [3.26], and together state that we have equal angles if and only if we have equal arcs. In [3.28], given equal chords, we must prove that we have equal arcs. In [3.29], given equal arcs, we must prove that we have equal chords. Hence, [3.29] is the converse of [3.28], and together state that we have equal chords if and only if we have equal arcs. Together, these propositions essentially state that we have equal chords if and only if we have equal angles if and only if the angles stand on equal arcs. PROPOSITION 3.26. ANGLES AND ARCS I. In equal circles, equal angles at the centers or on the circumferences stand upon arcs of equal length. PROOF. In equal circles (oACB, oDFE), equal angles at the centers (ZAOB, ZDHE) or at the circumferences (ZACB, ZDFE) stand on equal arcs. c F .\u25a0 Figure 3.2.30. [3.26] We prove each claim separately: 1. Suppose that ZAOB = ZDHE. We wish to show that these angles stand on equal arcs. 3.2. PROPOSITIONS FROM BOOK III 161 Because oACB, oDFE are equal, their radii are equal [Def. 3.1]. There- fore, the two triangles AAOB, ADHE have the sides AO, OB in one respec- tively equal to the sides DH, HE in the other, and ZAOB = ZDHE by hy- pothesis. By [1.4], AAOB ^ ADHE, and so AB = DE. Again, since the angles ZACB, ZDFE are the halves of the equal angles ZAOB, ZDHE [3.20], ZACB = ZDFE. By [Def. 3.10], ACB ~ DEE, and their chords AB, DE have been proved equal. By [3.24], the segments AB, DE are equal. And taking these equals from the circles which are equal by hypothesis, we have that the remaining arcs AGB, DKE are equal. 2. Now suppose that ZACB = ZDFE. Since ZACB = 2 \u2022 ZAOB and ZDFE = 2 \u2022 ZDHE [3.20], the proof follows from part 1. \u25a1 COROLLARY. 1. If the opposite angles of a cyclic quadrilateral are equal, one of its diagonals must be a diameter of the circumscribed circle. COROLLARY. 2. Parallel chords in a circle intercept equal arcs. COROLLARY. 3. If two chords intersect at any point within a circle, the sum of the opposite arcs which they intercept is equal to the arc which parallel chords intersecting on the circumference intercept. If two chords intersect at any point outside a circle, the difference of the arcs they intercept is equal to the arc which parallel chords intersecting on the circumference intercept. COROLLARY. 4. If two chords intersect at right angles, the sum of the oppo- site arcs which they intercept on the circle is a semicircle. PROPOSITION 3.27. ANGLES AND ARCS II. In equal circles, angles at the centers or at the circumferences which stand on equal arcs are equal in measure. PROOF. In equal circles (oACB, oDFE), angles at the centers (ZAOB, ZDHE) or at the circumferences (ZACB, ZDFE) which stand on equal arcs (AB, DE) are equal in measure. 3.2. PROPOSITIONS FROM BOOK III 162 b I Figure 3.2.31. [3.27] We prove each claim separately: 1. Consider the angles at the centers (ZAOB, ZDHE). Suppose that ZAOB > ZDHE and that ZAOL = ZDHE. Since the circles are equal, the arc AL is equal to arc DE [3.26]. But AB = DE by hypothesis. Hence AB = AL and AB = AL + LB, a contradiction. A corresponding contradiction follows if we assume that ZAOB ZACE. But ZACB is a right angle by part 1 of the proof, and so ZACE is acute. 3. Consider the arc ACD is less than a semicircle; similarly to the proof of part 2, we obtain that ZACD is obtuse. \u25a1 COROLLARY. 1. If a parallelogram is inscribed in a circle, its diagonals intersect at the center of the circle. COROLLARY. 2. [3.31] holds if arcs are replaced by chords of appropriate length, mutatis mutandis. Proposition 3.32. TANGENT-CHORD ANGLES RELATED TO ANGLES ON THE CIRCUMFERENCE WHICH STAND ON THE CHORD. If a line is tangent to a circle, and from the point of intersection a chord is constructed cut- ting the circle, the angles made by this chord with the tangent are respectively equal to the angles in the alternate arcs of the circle. PROOF. If a line (EF) is a tangent to a circle, and from the point of in- tersection (A) a chord (AC) is constructed cutting the circle, we claim that the angles made by this chord with the tangent are respectively equal to the angles in the alternate arcs of the circle. We shall prove this in two cases. A FIGURE 3.2.36. [3.321(a) 1. Construct the figures from Fig. 3.2.36(a). We wish to show that ZABC = ZFAC. 3.2. PROPOSITIONS FROM BOOK III 167 If AC passes through the center of oABC, then the proposition is evident because the angles are right angles. Otherwise, from the point of intersection A construct AB such that AB _L EF. Join BC. Because EF is tangent to the circle and AB is constructed from the point of intersection and is perpendicular to EF, AB passes through the center of oABC [3.19]. Therefore, ZACB is right [3.31], and the sum of the two remaining angles \/ABC + \/CAB equals one right angle. Since the angle \/BAF is right by construction, we have that \/ABC + \/CAB = \/BAF. From this we obtain \/ABC = \/BAF – \/CAB = \/CAF. 2. Construct the figures from Fig. 3.2.36(\/3). =. Figure 3.2.37. [3.32K\/3) Take any point D on the arc AC. We wish to prove that \/CAE = AC DA. Since the quadrilateral ABCD is cyclic, the sum of the opposite angles \/ABC + \/LCD A equals two right angles [3.22] and is therefore equal to the sum \/FAC + ZCAE. However, \/ABC = ZFAC by part 1. Removing them, we obtain \/LCD A = \/CAE. \u25a1 Alternatively: PROOF. Construct the figures from Fig. 3.2.36(\/3). Take any point G in the semicircle AGB. Join AG, GB, GC. Then we have that \/AGB = \/FAB, since each angle is right, and \/CGB = \/CAB [3.21]. Therefore \/AGC = \/FAC. Again, join BD, CD. Then \/BDA = \/BAE, since each angle is right, and \/CDB = \/CAB [3.21]. Hence, \/CD A = \/CAE. \u25a1 Exercises. 1. If two circles touch, any line constructed through the point of intersec- tion will cut off similar segments. 3.2. PROPOSITIONS FROM BOOK III 168 2. If two circles touch and any two lines are constructed through the point of intersection (cutting both circles again), the chord connecting their points of intersection with one circle is parallel to the chord connecting their points of intersection with the other circle. 3. Suppose that ACB is an arc of a circle, CE a tangent at C (meeting the chord AB extended to E), and AD JL AB where D is a point of AB. Prove that if BE be bisected at C then the arc AC = 2 \u2022 CB. 4. If two circles touch at a point A and if ABC is a chord through A, meeting the circles at points B and C, prove that the tangents at B and C are parallel to each other, and that when one circle is within the other, the tangent at B meets the outer circle at two points equidistant from C. 5. If two circles touch externally, their common tangent at either side stands opposite a right angle at the point of intersection, and its square is equal to the rectangle contained by their diameters. Proposition 3.33. CONSTRUCTING ARCS GIVEN A SEGMENT AND AN ANGLE. It is possible to construct an arc of a circle on a given segment such that the arc contains an angle equal to a given angle. PROOF. We wish to construct an arc (AB) of a circle (oABC) on a given segment (AB) which contains an angle equal to a given angle (ZFGH). \u25a0r Figure 3.2.38. [3.33] If ZFGH is a right angle, construct a semicircle on the given line. The proof follows since the angle in a semicircle is a right angle. 3.2. PROPOSITIONS FROM BOOK III 169 Otherwise, on the given segment AB, construct the angle ZBAE such that ZBAE = ZFGH. Construct AC _L AE and BC _L AB. With AC as diameter, construct the circle oCBA. We claim that oCBA is the required circle. The circumference of oCBA contains the point B (since ZABC is right [3.31]) and also touches AE (since ZCAE is right [3.16]). Therefore, ZBAE is equal to the angle in the alternate segment [3.32]; that is, ZBAE = ZACB. But ZBAE = ZFGH by construction, and so ZFGH = ZACB, and the arc AB contains the angle ZACB. \u25a1 Exercises. 1. Construct a triangle, being given the base, vertical angle, and any of the following data: (a) a perpendicular. (b) the sum or difference of the sides. (c) the sum or difference of the squares of the sides. (d) the side of the inscribed square on the base. (e) the median that bisects the base. 2. If lines are constructed from a fixed point to all the points of the circum- ference of a given circle, prove that the locus of all their points of bisection is a circle. 3. Given the base and vertical angle of a triangle, find the locus of the midpoint of the line joining the vertices of equilateral triangles constructed on the sides. 4. In the same case, find the loci of the angular points of a square con- structed on one of the sides. Proposition 3.34. DIVIDING ARCS GIVEN AN ANGLE. It is possible to divide the circumference of a circle such that the separated arc contains an angle equal to a given angle. PROOF. From a given circle (oBCA), we wish to cut off an arc which con- tains an angle equal to a given angle (ZFGH). 3.2. PROPOSITIONS FROM BOOK III 170 i. H Figure 3.2.39. [3.34] Take any point A on the circumference and construct the tangent AD. Con- struct the angle ZD AC such that ZD AC = ZFGH. Take any point B on the al- ternate arc. Join BA, BC. Then ZD AC = ZABC [3.32]. But ZD AC = ZFGH by construction, and so ZABC = ZFGH. \u25a1 PROPOSITION 3.35. AREAS OF RECTANGLES CONSTRUCTED ON CHORDS. If two chords of a circle intersect at one and only one point within the circle, then the area of the rectangle contained by the divided segments of the first chord is equal in area to the rectangle contained by the divided segments of the second chord [Def. 2.4]. PROOF. If two chords (AB, CD) of a circle (oACB) intersect at a point (E) within the circle, the rectangles (AE.EB, CE.ED) contained by the segments are equal in area. We prove this proposition in four cases: i: Figure 3.2.40. [3.35], case 1 1. If the point of intersection is the center of oACB, each rectangle is equal in area to the square of the radius. Hence, the proof. 3.2. PROPOSITIONS FROM BOOK III 171 Figure 3.2.41. [3.35], case 2 2. Suppose that the chord AB passes through the center of oABC and that the chord CD does not. Further suppose that AB _L CD. Join OC. Because AB passes through the center and cuts the other chord CD which does not pass through the center at right angles, AB bisects CD [3.3]. Because AB is divided equally at O and unequally at F, by [2.5], we have that AE.EB + OE 2 = OB 2 . Since OB = OC, we also have that AE.EB + OE 2 = OC 2 . But OC 2 = OE 2 + EC 2 [1.47]: therefore AE.EB + OE 2 = OE 2 + EC 2 . Subtracting OE 2 from both sides of the equality, we obtain AE.EB = EC 2 . But EC 2 = CE.ED since CE = ED. Therefore, AE.EB = CE.ED. Figure 3.2.42. [3.35], case 3 3. Let O be the center of oACB, and let AB pass through the center of oACB and cut CD such that AB \/ CD. Construct OF _L CD [1.11]. Join OC, OD. Since CD is cut at right angles by OF and OF passes through O, CD is bisected at F [3.3] and divided un- equally at E. Hence by [2.5], CE.ED + FF 2 = FD 2 . Adding OF 2 to each side of the equality, we obtain: CE.ED + F\u00a3 2 + OF 2 = FD 2 + OF 2 CE.ED + OF 2 = 0\u00a3> 2 CE.ED + OF 2 = 0\u00a3 2 3.2. PROPOSITIONS FROM BOOK III 172 Again, since AB is bisected at O and divided unequally at E, AE.EB + OE 2 = OB 2 [2.5]. It follows that CE.ED + OE 2 = AE.EB + 0\u00a3 2 C\u00a3.\u00a3L> = AE.EB 4. Suppose neither chord passes through the center. Through E, their point of intersection, construct the diameter FG. By case 3, the rectangle EE. EG = AE.EB and EE. EG = CE.ED. Hence, AE.EB = CE.ED. \u25a1 COROLLARY. 1. If a chord of a circle is divided at any point within the circle, the rectangle contained by its segments is equal to the difference between the square of the radius and the square of the segment constructed from the center to the point of section. COROLLARY. 2. If the rectangle contained by the segments of one of two intersecting segments is equal to the rectangle contained by the segments of the other, the four endpoints are coney die. COROLLARY. 3. If two triangles are equiangular, the rectangle contained by the non-corresponding sides about any two equal angles are equal. 3.2. PROPOSITIONS FROM BOOK III 173 Let A ABO, AD CO be the equiangular triangles, and let them be placed so that the equal angles at O are vertically opposite and that the non-corresponding sides AO, CO stand in one segment. Then the non-corresponding sides BO, OD form the segment BD. Since ZABD = ZACD, the points A, B, C, D are con- cyclic [3.21, Cor. 1], Hence, AO.OC = BO.OD [3.35]. Exercises. 1. In any triangle, the rectangle contained by two sides is equal in area to the rectangle contained by the perpendicular on the third side and the diame- ter of the circumscribed circle. Definition: The supplement of an arc is the difference between the arc and a semicircle. 2. The rectangle contained by the chord of an arc and the chord of its supplement is equal to the rectangle contained by the radius and the chord of twice the supplement. 3. If the base of a triangle is given with the sum of the sides, the rectangle contained by the perpendiculars from the endpoints of the base on the external bisector of the vertical angle is given. 4. If the base and the difference of the sides is given, the rectangle con- tained by the perpendiculars from the endpoints of the base on the internal bisector is given. 5. Through one of the points of intersection of two circles, construct a secant so that the rectangle contained by the intercepted chords may be given, or is a maximum. 6. If the sum of two arcs AC, CB of a circle is less than a semicircle, the rectangle AC.CB contained by their chords is equal to the rectangle contained by the radius and the excess of the chord of the supplement of their difference above the chord of the supplement of their sum. 3.2. PROPOSITIONS FROM BOOK III 174 Figure 3.2.45. [5.35, #6] Construct the diameter DE such that DE _L AB, and construct the chords CF, BG parallel to DE. It is evident that the difference between the arcs AC, CB is equal to 2- CD, and therefore equals CD + EF. Hence the arc CBF is the supplement of the difference and CF is the chord of that supplement. Again, since the angle ZABG is right, the arc ABG is a semicircle. Hence BG is the supplement of the sum of the arcs AC, CB, and therefore the segment BG is the chord of the supplement of the sum. By #1, the rectangle AC.CB is equal to the rectangle contained by the diameter and CI, and therefore equal to the rectangle contained by the radius and 2 \u2022 CI. But the difference between CF and BG is evidently equal to 2 \u2022 CI. Hence the rectangle AC.CB is equal to the rectangle contained by the radius and the difference between the chords CF, BG. 7. If we join AF, BF, we find that the rectangle AF.FB is equal in area to the rectangle contained by the radius and 2 \u2022 FI; that is, it is equal to the rectangle contained by the radius and the sum of CF and BG. Hence, if the sum of two arcs of a circle is greater than a semicircle, the rectangle contained by their chords is equal to the rectangle contained by the radius and the sum of the chords of the supplements of their sum and their difference. 8. Through a given point, construct a transversal cutting two given lines so that the rectangle contained by the segments intercepted between it and the line may be given. PROPOSITION 3.36. THE AREA OF RECTANGLES CONSTRUCTED ON A TANGENT AND A POINT OUTSIDE THE CIRCLE I. Suppose we are given a circle and a point outside of the circle. If two segments are constructed from the point to the circle, the first of which intersects the circle at two points and the 3.2. PROPOSITIONS FROM BOOK III 175 second of which is tangent to the circle, then the area of the rectangle contained by the subsegments of the first segment is equal to the square on the tangent. PROOF. If from any point (P) outside of the circle oATB two segments are constructed to meet P, one of which (PT) is a tangent and the other (PA) a secant, then the rectangle contained by the segments of the secant (AP.BP) is equal in area to the square of the tangent (PT); or, AP.BP = PT 2 . We solve the proposition in two cases: Figure 3.2.46. [3.36], case 1 1. Let PA pass through the center O of oATB. Join OT. Because AB is bisected at O and divided externally at P, the rectangle AP.BP + OB 2 = OP 2 [2.6]. Since PT is a tangent to oATB and OT is constructed from the center to the point of intersection, the angle ZOTP is right [3.18]. Hence OT 2 + PT 2 = OP 2 . Therefore AP.BP + OB 2 = OT 2 + PT 2 . But OB 2 = OT 2 , and so AP.BP = PT 2 . Figure 3.2.47. [3.36], case 2 2. If AB does not pass through the center O, construct the perpendicular OC on AB. Join OT, OB, OP. Because OC, a segment through the center, cuts AB, which does not pass through the center at right angles, OC bisects AB [3.3]. Since AB is bisected at C and divided externally at P, we have that AP.BP + CB 2 = CP 2 [2.6]. Adding OC 2 to each side, we obtain: 3.2. PROPOSITIONS FROM BOOK III 176 AP.BP + CB 2 + OC 2 = AP.BP + OP 2 = CP 2 + OC 2 OP 2 We also have that OT 2 + PT 2 = OP 2 , from which it follows that AP.BP + OP 2 = OT 2 + PT 2 . Subtracting OP 2 and OT 2 (since OB = OT), we have that Note: The two propositions [3.35] and [3.36] may be written as one state- ment: the rectangle AP.BP contained by the segments of any chord of a given circle passing through a fixed point P, either within or outside of the circle, is constant. Suppose O is the center the circle, and join OA, OB, OP. Then AOAB is an isosceles triangle, and OP is a segment constructed from its vertex to a point P in the base or the extended base. It follows that the rectangle AP.BP is equal to the difference of the squares of OB and OP; therefore, it is constant. COROLLARY. 1. If two segment AB, CD are extended to meet at P, and if the rectangle AP.BP = CP. DP, then the points A, B, C, D are concyclic (compare p3.35, Cor. 2]). COROLLARY. 2. Tangents to two circles from any point in their common chord are equal (compare [3.17, #6]). COROLLARY. 3. The common chords of any three intersecting circles are concurrent (compare [3.17, #7]). 1. If from the vertex A of AABC, the segment AD is constructed which meets CB extended to D and creates the angle ZBAD = ZACB, prove that DB.DC = DA 2 . Proposition 3.37. THE AREA OF RECTANGLES CONSTRUCTED ON A TANGENT AND A POINT OUTSIDE THE CIRCLE II. Suppose we are given a circle and a point outside of the circle. If two segments are constructed from the point to the circle, the first of which intersects the circle at two points, and the area of the rectangle contained by the subsegments of the first segment is equal to the square on the second segment, then the second segment is tangent to the circle. AP.BP = PT 2 . \u25a1 Exercise. 3.2. PROPOSITIONS FROM BOOK III 177 PROOF. If the rectangle (AP.BP) contained by the segments of a secant and constructed from any point (P) outside of the circle (oATB) is equal in area to the square on the segment (PT) constructed from the same point to meet the circle, then the segment which meets the circle is a tangent to that circle. Figure 3.2.48. [3.37] From P, construct PQ touching the circle [3.17]. Let O be the center of oATQ and join OP, OQ, OT. By hypothesis, AP.BP = PT 2 ; by [3.36], AP.BP = PQ 2 . Hence PT 2 = PQ 2 , and so PT = PQ. Consider the triangles AOTP, AOQP. Each have OT = OQ, TP = QP, and the base OP in common. By [1.8], AOTP ^ AOQP, and so AOTP = AOQP. But AOQP is a right angle since PQ is a tangent [3.38]; hence AOTP is right, and therefore PT is a tangent to oATB [3.16]. \u25a1 COROLLARY. 1. Suppose we are given a circle and a point outside of the circle where two segments are constructed from the point to the circle, the first of which intersects the circle at two points. Then the second segment is tangent to the circle if and only if the area of the rectangle contained by the subsegments of the first segment is equal to the square on the tangent. Exercises. 1. Construct a circle passing through two given points and fulfilling either of the following conditions: (a) touching a given line; (b) touching a given circle. 2. Construct a circle through a given point and touching two given lines; or touching a given line and a given circle. 3. Construct a circle passing through a given point having its center on a given line and touching a given circle. 3.2. PROPOSITIONS FROM BOOK III 178 4. Construct a circle through two given points and intercepting a given arc on a given circle. 5. If A, B, C, D are four collinear points and EF is a common tangent to the circles constructed upon AB, CD as diameters, then prove that the triangles AAEB, ACFD are equiangular. 6. The diameter of the circle inscribed in a right-angled triangle is equal to half the sum of the diameters of the circles touching the hypotenuse, the per- pendicular from the right angle of the hypotenuse, and the circle constructed about the right-angled triangle. Examination questions on chapter 3 1. What is the subject-matter of chapter 3? 2. Define equal circles. 3. Define a chord. 4. When does a secant become a tangent? 5. What is the difference between an arc and a sector? 6. What is meant by an angle in a segment? 7. If an arc of a circle is one-sixth of the whole circumference, what is the magnitude of the angle in it? 8. What are segments? 9. What is meant by an angle standing on a segment? 10. What are concyclic points? 11. What is a cyclic quadrilateral? 12. How many intersections can a line and a circle have? 13. How many points of intersection can two circles have? 14. Why is it that if two circles touch they cannot have any other common point? 15. State a proposition that encompasses [3.11] and [3.12]. 16. What proposition is #16 a limiting case of? 17. What is the modern definition of an angle? 18. How does the modern definition of an angle differ from Euclid’s? 19. State the relations between [3.16], [3.18] and [3.19]. 20. What propositions are [3.16], [3.18] and [3.19] limiting cases of? 21. How many common tangents can two circles have? 22. What is the magnitude of the rectangle of the segments of a chord constructed through a point 3.65m distant from the center of a circle whose radius is 4.25m? 3.2. PROPOSITIONS FROM BOOK III 179 23. The radii of two circles are 4.25 and 1.75 ft respectively, and the dis- tance between their centers 6.5 ft. Find the lengths of their direct and their transverse common tangents. 24. If a point is h feet outside the circumference of a circle whose diameter is 7920 miles, prove that the length of the tangent constructed from it to the circle is ^Zhjl miles. 25. Two parallel chords of a circle are 12 inches and 16 inches respectively and the distance between them is 2 inches. Find the length of the diameter. 26. What is the locus of the centers of all circles touching a given circle in a given point? 27. What is the condition that must be fulfilled that four points may be concyclic? 28. If the angle in a segment of a circle equals 1.5 right angles, what part of the whole circumference is it? 29. Mention the converse propositions of chapter 3 which are proved di- rectly. 30. What is the locus of the midpoints of equal chords in a circle? 31. The radii of two circles are 6 and 8, and the distance between their centers is 10. Find the length of their common chord. 32. If a figure of any even number of sides is inscribed in a circle, prove that the sum of one set of alternate angles is equal to the sum of the remaining angles. Chapter 3 exercises. 1. If two chords of a circle intersect at right angles, the sum of the squares on their segments is equal to the square on the diameter. 2. If a chord of a given circle stands opposite a right angle at a fixed point, the rectangle of the perpendiculars on it from the fixed point and from the center of the given circle is constant. Also, the sum of the squares of perpen- diculars on it from two other fixed points (which may be found) is constant. 3. If through either of the points of intersection of two equal circles any line is constructed meeting them again in two points, these points are equally distant from the other intersection of the circles. 4. Construct a tangent to a given circle so that the triangle formed by it and two fixed tangents to the circle shall be: (a) a maximum; (b) a minimum. 3.2. PROPOSITIONS FROM BOOK III 180 5. If through the points of intersection A, B of two circles any two segments ACD, BEF are constructed parallel to each other which meet the circles again at C, D, E, F, then we have that CD = EF. 6. In every triangle, the bisector of the greatest angle is the least of the three bisectors of the angles. 7. The circles whose diameters are the four sides of any cyclic quadrilateral intersect again in four concyclic points. 8. The four angular points of a cyclic quadrilateral determine four triangles whose orthocenters (the intersections of their perpendiculars) form an equal quadrilateral. 9. If through one of the points of intersection of two circles we construct two common chords, the segments joining the endpoints of these chords make a given angle with each other. 10. The square on the perpendicular from any point in the circumference of a circle on the chord of contact of two tangents is equal to the rectangle of the perpendiculars from the same point on the tangents. 11. Find a point on the circumference of a given circle such that the sum of the squares on whose distances from two given points is either a maximum or a minimum. 12. Four circles are constructed on the sides of a quadrilateral as diame- ters. Prove that the common chord of any two on adjacent sides is parallel to the common chord of the remaining two. 13. The rectangle contained by the perpendiculars from any point in a circle on the diagonals of an inscribed quadrilateral is equal to the rectangle contained by the perpendiculars from the same point on either pair of opposite sides. 14. The rectangle contained by the sides of a triangle is greater than the square on the internal bisector of the vertical angle by the rectangle contained by the segments of the base. 15. If through A, one of the points of intersection of two circles, we con- struct any line ABC which cuts the circles again at B and C, the tangents at B and C intersect at a given angle. 16. If a chord of a given circle passes through a given point, the locus of the intersection of tangents at its endpoints is a straight line. 17. The rectangle contained by the distances of the point where the in- ternal bisector of the vertical angle meets the base and the point where the perpendicular from the vertex meets it from the midpoint of the base is equal to the square on half the difference of the sides. 3.2. PROPOSITIONS FROM BOOK III 181 18. State and prove the proposition analogous to [3.17] for the external bisector of the vertical angle. 19. The square on the external diagonal of a cyclic quadrilateral is equal to the sum of the squares on the tangents from its endpoints to the circumscribed circle. 20. If a “movable” circle touches a given circle and a given line, the chord of contact passes through a given point. 21. If A, B, C are three points in the circumference of a circle, and D, E are the midpoints of the arcs AB, AC, and if the segment DE intersects the chords AB, AC at F and G, then AF = AG. 22. If a cyclic quadrilateral is such that a circle can be inscribed in it, the lines joining the points of contact are perpendicular to each other. 23. If through the point of intersection of the diagonals of a cyclic quadri- lateral the minimum chord is constructed, that point will bisect the part of the chord between the opposite sides of the quadrilateral. 24. Given the base of a triangle, the vertical angle, and either the internal or the external bisector at the vertical angle, construct the triangle. 25. If through the midpoint A of a given arc BAC we construct any chord AD, cutting BC at E, then the rectangle AD.AE is constant. 26. The four circles circumscribing the four triangles formed by any four lines pass through a common point. 27. If X, Y, Z are any three points on the three sides of a triangle AABC, the three circles about the triangles AYAZ, AZBX, AXCY pass through a common point. 28. If the position of the common point in the previous exercise are given, the three angles of the triangle AXYZ are given, and conversely. 29. Place a given triangle so that its three sides shall pass through three given points. 30. Place a given triangle so that its three vertices shall lie on three given lines. 31. Construct the largest triangle equiangular to a given one whose sides shall pass through three given points. 32. Construct the smallest possible triangle equiangular to a given one whose vertices shall lie on three given lines. 33. Construct the largest possible triangle equiangular to a given triangle whose sides shall touch three given circles. 34. If two sides of a given triangle pass through fixed points, the third touches a fixed circle. 3.2. PROPOSITIONS FROM BOOK III 182 35. If two sides of a given triangle touch fixed circles, the third touches a fixed circle. 36. Construct an equilateral triangle having its vertex at a given point and the endpoints of its base on a given circle. 37. Construct an equilateral triangle having its vertex at a given point and the endpoints of its base on two given circles. 38. Place a given triangle so that its three sides touch three given circles. 39. Circumscribe a square about a given quadrilateral. 40. Inscribe a square in a given quadrilateral. 41. Construct the following circles: (a) orthogonal (cutting at right angles) to a given circle and passing through two given points; (b) orthogonal to two others, and passing through a given point; (c) orthogonal to three others. 42. If from the endpoints of a diameter AB of a semicircle two chords AD, BE are constructed which meet at C, we have that ACAD + BC.BE = AB 2 . 43. If ABCD is a cyclic quadrilateral, and if we construct any circle passing through the points A and B, another through B and C, a third through C and D, and a fourth through D and A, then these circles intersect successively at four other points E, F, G, H, forming another cyclic quadrilateral. 44. If AABC is an equilateral triangle, what is the locus of the point M, if MA = MB + MCI 45. In a triangle, given the sum or the difference of two sides and the angle formed by these sides both in magnitude and position, the locus of the center of the circumscribed circle is a straight line. 46. Construct a circle: (a) through two given points which bisect the circumference of a given cir- cle; (b) through one given point which bisects the circumference of two given circles. 47. Find the locus of the center of a circle which bisects the circumferences of two given circles. 48. Construct a circle which bisects the circumferences of three given cir- cles. 49. If CD is a perpendicular from any point C in a semicircle on the diam- eter AB, oEFG is a circle touching DB at E, CD at F, and the semicircle at G, then prove that: (a) the points A, F, G are collinear; (b) AC = AE. 3.2. PROPOSITIONS FROM BOOK III 183 50. Being given an obtuse-angled triangle, construct from the obtuse angle to the opposite side a segment whose square is equal to the rectangle contained by the segments into which it divides the opposite side. 51. If O is a point outside a circle whose center is E and two perpendicular segments passing through O intercept chords AB, CD on the circle, then prove that AB 2 + CD 2 + 40E 2 = SR 2 . 52. The sum of the squares on the sides of a triangle is equal to twice the sum of the rectangles contained by each perpendicular and the portion of it comprised between the corresponding vertex and the orthocenter. It is also equal to 12R 2 minus the sum of the squares of the distances of the orthocenter from the vertices. 53. If two circles touch at C, if D is any point outside the circles at which their radii through C stands opposite equal angles, and if DE, DF are tangent from D, prove that DE.DF = DC 2 . CHAPTER 4 Inscription and Circumscription This chapter contains sixteen propositions, four of which relate to trian- gles, four to squares, four to pentagons, and four to miscellaneous figures. 4.1. Definitions 1. If two polygons are related such that the angular points of one lie on the sides of the other, then: (a) the former is said to be inscribed in the latter; (b) the latter is said to be circumscribed around or about the former. 2. A polygon is said to be inscribed in a circle when its angular points are on the circumference. Reciprocally, a polygon is said to be circumscribed about or around a circle when each side touches the circle. 3. A circle is said to be inscribed in a polygon when it touches each side of the figure. Reciprocally, a circle is said to be circumscribed about or around a polygon when it passes through each angular point of the figure. FIGURE 4.1.1. [Def. 4.1] The hexagon ABCDEF is inscribed in the circle oABC; the circle oABC is circumscribed about the hexagon ABCDEF. 4. A polygon which is both equilateral and equiangular is said to be regu- lar. 4.2. Propositions from Book IV Proposition 4.1. CONSTRUCTING A CHORD INSIDE A CIRCLE. In a given circle, it is possible to construct a chord equal in length to a given segment not greater than the circle’s diameter. 184 4.2. PROPOSITIONS FROM BOOK IV 185 PROOF. Given with diameter AC, we wish to construct a chord in oABC equal in length to a given segment, DG (where AC > DG). Figure 4.2.1. [4.1] If AC = DG, then the required chord already exists within the circle (i.e., its diameter). Otherwise, cut subsegment AE from diameter AC such that AE = DG [1.3]. With A as center and AE as radius, construct the circle oEBF, cutting the circle oABC at the points B, F. Join AB. We claim that AB is the required chord. Because A is the center of oEBF, AB = AE. But AE = DG by construc- tion, and so AB = DG. Since AB is a chord of oABC, the proof follows. \u25a1 PROPOSITION 4.2. INSCRIBE A TRIANGLE INSIDE A CIRCLE. In a given circle, it is possible to inscribe a triangle equiangular to a given trian- gle. PROOF. We wish to inscribe a triangle equiangular to a given triangle (ADEF) in a given circle (oABC). A H Figure 4.2.2. [4.2] At a point A on the circumference of oABC, construct the tangent line GAR. Construct AH AC = ZDEF and ZGAB = ZDFE [1.23]. Join BC. We claim that AABC fulfills the required conditions. 4.2. PROPOSITIONS FROM BOOK IV 186 Since ZDEF = ZHAC by construction and ZHAC = ZABC in the al- ternate segment [3.32], we have that ZDEF = ZABC. Similarly, ZDFE = ZACB. By [1.32], ZFDE = ZBAC. Hence the triangle AABC inscribed in oABC is equiangular to ADEF. \u25a1 Proposition 4.3. CIRCUMSCRIBE A TRIANGLE ABOUT A CIRCLE. It is possible to circumscribe a triangle about a circle such that the triangle is equiangular to a given triangle. PROOF. We wish to to construct a triangle equiangular to a given triangle (ADEF) about a given circle (oABC). IT. H Figure 4.2.3. [4.3] Extend side DE of ADEF to the segment GH, and from the center O of oBCA construct any radius OA. Construct ZAOB = ZGEF and ZAOC = ZHDF [1.23]. At the points A, B, and C, construct the tangents LM, MN, NL to oBCA. We claim that ALMN fulfills the required conditions. Because AM touches oBCA at A, the angle ZOAM is right [3.18]. Simi- larly, the angle ZMBO is right; but the sum of the four angles of the quadri- lateral OAMB is equal to four right angles [1.32, Cor.3]. Therefore the sum of the two remaining angles ZAOB + ZAMB is two right angles. By [1.13], ZGEF + ZFED is two right angles [1.13], and so ZAOB + ZAMB = ZGEF + ZFED. But ZAOB = ZGEF by construction; hence ZAMB = ZFED; simi- larly, ZALC = ZEDF. Therefore by [1.32], ZBNC = ZDFE, and the triangle ALMN is equiangular to ADEF. \u25a1 Proposition 4.4. INSCRIBE A CIRCLE IN A TRIANGLE. It is possible to inscribe a circle in a given triangle. 4.2. PROPOSITIONS FROM BOOK IV 187 PROOF. We wish to inscribe a circle (oDEF) in a given triangle (AABC). Figure 4.2.4. [4.4] Bisect angles ZCAB, ZABC of AABC by the segments AO, BO; we claim that O, their point of intersection, is the center of the required circle. From O construct the perpendicular segments OD, OE, OF on the sides of AABC. Consider the triangles AOAE and AOAF: ZOAE = ZOAF by construction; ZAEO = ZAFO because each is right; and the side OA stands in common. Hence, OE = OF [1.26]. Similarly, OD = OF. Therefore OD = OE = OF, and by [3.9], the circle constructed with O as center and OD as radius will pass through the points D, E, F by construction. Since each of the angles ZODB, ZOEA, ZOFA is right, each touches the respective sides of the triangle AABC [3.16]. Therefore, the circle oDEF is inscribed in the triangle AABC. \u25a1 Definition: The bisectors of the three internal angles of a triangle are concurrent. Their point of intersection is called the incenter of the triangle. Exercises. 1. In [4.4]: if the points O, C are joined, prove that the angle ZACB is bisected. Hence, we prove the existence of the incenter of a triangle. 2. If the sides BC, CA, AB of the triangle AABC are denoted by a, b, c, and half their sum is denoted by s, the distances of the vertices A, B, C of the triangle from the points of contact of the inscribed circle are respectively s-a, s-b, s-c. 3. If the external angles of the triangle AABC are bisected as in Fig. 4.2.5, the three angular points O’, O”, O'” of the triangle formed by the three bisectors are the centers of three circles, each touching one side externally and the other two when extended. These three circles are called the escribed circles of the triangle A ABC. 4. The distances of the vertices A, B, C from the points of contact of the escribed circle which touches AB externally are s-b, s-a, s. 5. The center of the inscribed circle, the center of each escribed circle, and two of the angular points of the triangle are concyclic. Also, any two of the escribed centers are concyclic with the corresponding two of the angular points of the triangle. 6. Of the four points O, O’, O”, O”‘ ‘, any one is the orthocenter of the triangle formed by the remaining three. 7. The three triangles ABCO, ACAO,AABO are equiangular. 8. Confirm that CO. CO = ab, AO. AO = be, BO. BO = ca. 9. Since the whole triangle AABC is made up of the three triangles AAOB, ABOC, ACOA, we see that the rectangle contained by the sum of the three sides and the radius of the inscribed circle is equal to twice the area of the triangle. Hence, if r denotes the radius of the inscribed circle, rs =area of AABC. 4.2. PROPOSITIONS FROM BOOK IV 189 10. If r’ denotes the radius of the escribed circle which touches the side a externally, it may be shown that r'(s-a) =area of the triangle. 11. Show that rr’ = s-b.s-c. 12. Show that the square of the area = s.(s-a).(s-b).(s-c). 13. Show that the square of the area = r.r’ .r” .r'” . 14. If AABC is a right triangle where the angle at C is right, then r = s-c, r’ = s-b, r” = s-a, and r'” = s. 15. Given the base of a triangle, the vertical angle, and the radius of the inscribed or any of the escribed circles, construct it. Proposition 4.5. CIRCUMSCRIBE A CIRCLE ABOUT A TRIANGLE. It is possible to circumscribe a circle about a given triangle. PROOF. We wish to construct a circle (oABC) about a given triangle ( AABC). Figure 4.2.6. [4.5] Bisect sides BC, AC of AABC at the points D, E, respectively. Construct DO, EO at right angles to BC, CA. We claim that O, the point of intersection of the perpendiculars, is the center of the required circle oABC. Join OA, OB, OC. Consider the triangles ABDO, ACDO: they have the side BD = CD by construction, side DO in common, and ZBDO = ZCDO because each is right. By [1.4], ABDO ^ ACDO, and so BO = OC. Similarly, AO = OC. Therefore, AO = BO = CO, and by [3.9], the circle oABC con- structed with O as center and OA as radius will pass through the points A, B, and C; thus, oABC is circumscribed about the triangle AABC. \u25a1 4.2. PROPOSITIONS FROM BOOK IV 190 COROLLARY. 1. Since the perpendicular from O to AB bisects AB [3.3], we see that the perpendiculars at the midpoints of the sides of a triangle are concurrent. (See the following Definition.) Definition: The circle oABC is called the circumcircle, its radius the cir- cumradius, and its center the circumcenter of the triangle. Exercises. 1. The three altitudes of a triangle (AABC) are concurrent. (This proves the existence of the orthocenter of a circle.) 2. Prove that the three rectangles OA.OP, OB.OQ, OC.OR are equal in area. Figure 4.2.7. [4.5, #2] Definition: The circle with O as center, the square of whose radius is equal OA.OP = OB.OQ = OC.OR, is called the polar circle of the triangle AABC. 3. If the altitudes of a triangle are extended to meet a circumscribed circle, the intercepts between the orthocenter and the circle are bisected by the sides of the triangle. Definition: The nine-points circle is a circle that can be constructed for any given triangle. It is so named because it passes through nine significant concyclic points defined from the triangle. These nine points are: 4.2. PROPOSITIONS FROM BOOK IV 191 (a) the midpoint of each side of the triangle (b) the foot of each altitude (c) the midpoint of the line segment from each vertex of the triangle to the orthocenter (where the three altitudes meet; these line segments lie on their respective altitudes). [] See Fig. 4.2.9. 205_home_c.aniel_Documents_Euclid_Project_images-book4_Fig8p2p9 . FIGURE 4.2.8. [4.5, #4] The nine-points circle 4. The circumcircle of a triangle is the “nine points circle” of each of the four triangles formed by joining the centers of the inscribed and escribed circles. 5. The distances between the vertices of a triangle and its orthocenter are respectively the doubles of the perpendiculars from the circumcenter on the sides. 6. The radius of the “nine points circle” of a triangle is equal to half its circumradius. Note: the orthocenter, centroid, and circumcenter of any triangle are collinear; the line they lie on is named the Euler lin^j PROPOSITION 4.6. INSCRIBE A SQUARE IN A CIRCLE. It is possible to to inscribe a square in a given circle. PROOF. We wish to inscribe a square (BABCD) in a given circle (oABCD). Figure 4.2.9. [4.6] https : \/\/en.wikipedia. org\/wiki\/Nine-point_circle ” https : \/\/en . wikipedia . org\/wiki\/Leonhard_Euler 4.2. PROPOSITIONS FROM BOOK IV 192 Construct any two diameters AC, BD such that AC JL BD. Join AB, BC, CD, DA. We claim that BABCD is the required square. Let O be the center of oABCD. Then the four angles at O are equal since they are right angles. Hence the arcs on which they stand are equal [3.26] and the four chords on which they stand are equal in length [3.29]. Therefore the figure BABCD is equilateral. Again, because AC is a diameter, the angle ZABC is right [3.31]. Similarly, the remaining angles are right. It follows that BABCD is a square inscribed in oABCD. B Proposition 4.7. CIRCUMSCRIBE A SQUARE ABOUT A CIRCLE. It is possible to circumscribe a square about a given circle. PROOF. We wish to construct a circle (oABC) about the square (BABCD). f H Figure 4.2.10. [4.7] Through the center O construct any two diameters at right angles to each other, and at the points A, B, C, D construct the tangential segments HE, EF, FG, GH. We claim that BEFGH is the required square. Since AE touches the circle at A, the angle ZEAO is right [3.18] and there- fore equal to ZBOC, which is right by construction. Hence AE \\ OB. Similarly, EB || AO. Since AO = OB, the figure AOBE is a lozenge. Since the angle ZAOB is right, BAOBE is a square. Similarly, each of the figures BBOFC, BODGC, BAH DO is a square, and so BEFGH is a square. \u25a1 COROLLARY. 1. The circumscribed square has double the area of the in- scribed square. 4.2. PROPOSITIONS FROM BOOK IV 193 Proposition 4.8. INSCRIBE A CIRCLE IN A SQUARE. It is possible to inscribe a circle in a given square. PROOF. We wish to inscribe a circle (oABC) in a given square {[HEFGH). Figure 4.2.11. [4.8] Bisect the two adjacent sides EH, EF at the points A, B, and through A, B construct the segments AC, BD which are respectively parallel to EF, EH. We claim that O, the point of intersection of these parallels, is the center of the required circle oABC. Because [I] AO BE is a parallelogram, its opposite sides are equal; therefore AO = EB. But EB is half the side of the given square, and so AO is equal to half the side of the given square. This is similarly true for each of the segments OB, OC, OD. Therefore we have that OA = OB = OC = OD. Since they are perpendicular to the sides of the given square, the circle constructed with O as center and OA as radius is inscribed in the square. \u25a1 Proposition 4.9. CIRCUMSCRIBE A CIRCLE ABOUT A GIVEN SQUARE. It is possible to circumscribe a circle about a given square. PROOF. We wish to construct a circle (oABC) about a given square (BABCD). 4.2. PROPOSITIONS FROM BOOK IV 194 Figure 4.2.12. [4.9] Construct perpendicular diagonals AC, BD intersecting at O. We claim that O is the center of the required circle oABC. Since AABC is an isosceles triangle and the angle ZABC is right, each of the other angles equals half a right angle; therefore ZBAO equals half a right angle. Similarly, ZABO equals half a right angle; hence ZBAO = ZABO. By [1.6], AO = OB. Similarly, OB = OC and OC = OD. Hence the circle with O as center and OA as radius intersects through the points B, C, D and is evidently constructed about the square BABCD. \u25a1 PROPOSITION 4.10. CONSTRUCTION OF AN ISOSCELES TRIANGLE WITH BASE ANGLES DOUBLE THE VERTICAL ANGLE. It is possible to construct an isosceles triangle such that each base angle is double the vertical angle. PROOF. We wish to construct an isosceles triangle (AABD) where ZADB = ZABD = 2 \u2022 ZDAB 4.2. PROPOSITIONS FROM BOOK IV 195 Take any segment AB and divide it at C such that the rectangle AB.BC = AC 2 [2.6]. With A as center and AB as radius, construct the circle oBDE and on its circumference construct the chord BD = AC [4.1]. Join AD. We claim that AABD fulfills the required conditions. Join CD. About the triangle AACD, construct the circle oACD [4.5]. Be- cause by construction the rectangle AB.BC = AC 2 and AC = BD, each by con- struction, we have that the rectangle AB.BC = BD 2 . By [3.32], BD touches the circle oACD. Hence the angle \/BDC is equal to the angle at A in the alternate segment (or, \/BDC = \/DAB) [3.32]. To each angle, add \/CD A, and we have that \/BDA = \/LCD A + \/-DAB. But the exterior angle \/BCD of the triangle AACD is equal to the sum of the angles \/CD A and \/CAD (or \/BCD = \/CD A + \/DAB). Hence \/BDA = \/BCD. Since AB = AD, we have that \/BDA = \/ABD, from which it follows that \/BCD = \/ABD = \/CBD By [1.6], BD = CD; but BD = AC by construction. Therefore AC = CD, and by [1.5], \/CD A = \/DAB. But since \/BDA = \/CD A + \/DAB, \/BDA = 2 \u2022 \/DAB. It follows that each of the base angles of the triangle AABD is double of the vertical angle. \u25a1 Exercises. 1. Prove that AACD is an isosceles triangle whose vertical angle is equal to three times each of the base angles. 4.2. PROPOSITIONS FROM BOOK IV 196 2. Prove that BD is the side of a regular decagon inscribed in the circle oBDE. 3. If DB, DE, EF are consecutive sides of a regular decagon inscribed in a circle, prove that BF-BD = radius of a circle. 4. If E is the second point of intersection of the circle oACD with oBDE, then DE = DB. If AE, BE, CE, DE are joined, each of the triangles AACE, AADE is congruent with AABD. 5. AC is the side of a regular pentagon inscribed in the circle oACD, and EB the side of a regular pentagon inscribed in the circle oBDE. 6. Since AACE is an isosceles triangle, EB 2 -EA 2 = AB.BC = BD 2 ; that is, the square of the side of a pentagon inscribed in a circle exceeds the square of the side of the decagon inscribed in the same circle by the square of the radius. Proposition 4.11. INSCRIBE A REGULAR PENTAGON IN A GIVEN CIRCLE. It is possible to inscribe a regular pentagon in a given circle. PROOF. We wish to inscribe a regular pentagon in a given circle (oABC). Figure 4.2.14. [4.11] Construct an isosceles triangle having each base angle double the vertical angle [4.10], and inscribe a triangle AABD equiangular to it in the given circle oABC. Bisect the angles ZDAB, ZABD by the segments AC, BE. Join EA, ED, DC, CB. We claim that the figure ABCDE is a regular pentagon. Since each of the base angles ZBAD, ZABD is double of the angle ZADB (or ZBAD = 2 \u2022 ZADB = ZABD) and the segments AC, BE bisect them, we have that ZBAC = ZCAD = ZADB = ZDBE = ZEBA Therefore the arcs on which they stand are equal, and the five chords AB, BC, CD, DE, EA are equal in length. Hence, the figure ABCDE is equilateral. 4.2. PROPOSITIONS FROM BOOK IV 197 Again, because the arcs AB, DE are equal, if we add the arc BCD to both, the arc ABCD is equal to the arc BCDE, and therefore the angles ZAED, ZBAE which stand on them are equal [3.27]. Similarly, it can be shown that all the angles are equal; therefore ABCDE is equiangular. Hence, it is a regular pentagon. \u25a1 Exercises. 1. The figure formed by the five diagonals of a regular pentagon is another regular pentagon. 2. If the alternate sides of a regular pentagon are extended to meet, the five points of meeting form another regular pentagon. 3. Every two consecutive diagonals of a regular pentagon divide each other in the extreme and mean ratio. 4. Being given a side of a regular pentagon, construct it. 5. Divide a right angle into five equal parts. Proposition 4.12. CIRCUMSCRIBE A REGULAR PENTAGON ABOUT A GIVEN CIRCLE. It is possible to circumscribe a regular pentagon about a given circle. PROOF. We wish to construct a regular pentagon about a given circle (oABC). Figure 4.2.15. [4.12] Let the five points A, B, C, D, E on the circumference oABC be the vertices of any inscribed regular pentagon; at these points, construct tangents FG, GH, HI, IJ, JF. We claim that FGHIJ is a circumscribed regular pentagon. Let O be the center of oABC. Join OE, OA, OB. Because the angles 1.0 AF, ZOEF of the quadrilateral AOEF are right angles [3.18], the sum of the two 4.2. PROPOSITIONS FROM BOOK IV 198 remaining angles ZAOE + ZAFE equals two right angles. Similarly, the sum ZAOB + ZAGB equals two right angles; hence ZAOE + ZAFE = ZAOB + ZAGB But we have that ZAOE = ZAOB because they stand on equal arcs AE, AB [3.27]. Hence ZAFE = ZAGB. Similarly, the remaining angles of the figure FGHIJ are equal, and so FGHIJ is equiangular. Again, join OF, OG. Notice that the triangles AEOF, AAOF share equal sides AF, FE [3.17, #1], have side FO in common, and have equal bases AO, EO. Hence AEOF ^ AAOF, and so ZAFO = ZEFO [1.8]. Therefore ZAFO = ZAFE. Similarly, ZAGO = ZAGB. But ZAFE = ZAGB, and so ZAFO = ZAGO; also, ZFAO = ZGAO since each are right angles, and the two triangles AFAO, AG AO share side AO in common. By [1.26], AF = AG, and so GF = 2 \u2022 AF; similarly, JF = 2 \u2022 EF. And since AF = EF, GF = JF. Similarly, the remaining sides are equal; therefore the figure FGHIJ is equilateral and equiangular. Hence, it is a regular pentagon. \u25a1 Note: This proposition is a particular case of the following general theorem (which has a analogous proof): “If tangents are constructed to a circle at the angular points of an inscribed polygon of any number of sides, they will form a regular polygon of the same number of sides circumscribed to the circle.” Proposition 4.13. INSCRIBING A CIRCLE IN A REGULAR PENTA- GON It is possible to inscribe a circle in a regular pentagon. PROOF. We wish to inscribe a circle (oJFG) in a regular pentagon (ABCDE). c H Figure 4.2.16. [4.13] 4.2. PROPOSITIONS FROM BOOK IV 199 Bisect two adjacent angles ZJAF, ZFBG by the segments AO, BO; we claim that the point of intersection of the bisectors, O, is the center of the required circle oJFG. Join CO, and construct perpendiculars from O on the five sides of the pen- tagon. Notice that the triangles AABO, ACBO have the sides AB = BC by hypothesis, side BO in common, and ZABO = ZCBO by construction. By [1.4], ZBAO = ZBCO; however, ZBAO = ZBAE by construction. Therefore ZBCO = -ZBAE = -ZBCD 2 2 and hence CO bisects the angle ZBCD. Similarly, it may be proved that DO bisects the angle ZHDI and EO bi- sects the angle ZIEJ. Again, the triangles ABOF, ABOG have the angle ZOFA = ZOGB since each are right, ZOBF = OBG because OB bisects ZABC by construction, and they share side OB in common. Hence, OF = OG [1.26]. Similarly, all the perpendiculars from O on the sides of the pentagon are equal. Hence the circle whose center is O with radius OF touches all the sides of the pentagon and will therefore be inscribed in it; or, a circle may be in- scribed in any regular polygon. \u25a1 Proposition 4.14. CIRCUMSCRIBE A CIRCLE ABOUT A REGULAR PENTAGON. It is possible to circumscribe a circle about a regular pentagon. PROOF. We wish to construct a circle (oAED) about a regular pentagon (ABCDE). Figure 4.2.17. [4.14] 4.2. PROPOSITIONS FROM BOOK IV 200 Bisect two adjacent angles ZBAE, ZABC by the segments AO, BO. We claim that O, the point of intersection of the bisectors, is the center of the required circle oABC. Join OC, OD, OE. Then the triangles AABO, ACBO have the side AB = BC by hypothesis, BO common, and ZABO = ZCBO by construction. By [1.4], LB AO = ZBCO. But ZBAE = ZBCD by hypothesis. Since ZBAO = ZBAE by construction, ZBCO = ZBCD. Hence CO bisects the angle ZBCD. Similarly, it may be proved that DO bisects ZCDE and EO bisects the angle ZDEA. Again, because ZEAB \u2014 ZABC, their halves are equal, and ZOAB = ZOBA. By [1.4], OA = OB. Similarly, the segments OC, OD, OE are equal to one another and to OA. Therefore the circle constructed with O as center and OA as radius (oAED) passes through the points B, C, D, E and is constructed about the pentagon. \u25a1 Note: [4.13] and [4.14] are particular cases of the following theorem: “A regular polygon of any number of sides has one circle inscribed in it and an- other constructed about it, and both circles are concentric. Proposition 4.15. INSCRIBE A REGULAR HEXAGON IN A CIRCLE. It is possible to inscribe a regular hexagon in a circle. PROOF. We wish to to inscribe a regular hexagon (ABCDEF) in a given circle (oABC). Figure 4.2.18. [4.15] Take any point A on the circumference and join it to O, the center of the circle. Then with A as center and AO as radius, construct the circle oOBF, intersecting oABC at the points B, F. Join OB, OF and extend AO, BO, FO to meet oABC again at the points D, E, C. Join AB, BC, CD, DE, EF, FA; we claim that ABCDEF is the required hexagon. 4.2. PROPOSITIONS FROM BOOK IV 201 Each of the triangles AAOB, AAOF is equilateral. Hence the angles ZAOB, ZAOF are each one-third of two right angles; therefore ZEOF is one- third of two right angles. Again, the angles ZBOC, ZCOD, ZDOE are re- spectively equal to the angles ZEOF, ZFOA, ZAOB [1.15]. Therefore the six angles at the center are equal, because each is one-third of two right angles, and so by [3.29], AB = BC = CD = DE = EF = FA Hence the hexagon is equilateral. Since the arc AF = ED, to each arc add arc ABCD; it follows that the whole arc FABCD = ABODE, and therefore the angles ZDEF, ZEFA which stand on these arcs are equal [3.27]. Similarly, it may be shown that the other angles of the hexagon are equal. Hence ABCDE is equiangular and is there- fore a regular hexagon inscribed in the circle. \u25a1 COROLLARY. 1. The length of the side of a regular hexagon inscribed in a circle is equal to the circle’s radius. COROLLARY. 2. If three alternate angles of a hexagon are joined, they form an inscribed equilateral triangle. Exercises. 1. The area of a regular hexagon inscribed in a circle is equal to twice the area of an equilateral triangle inscribed in the circle. Also, the square of the side of the triangle equals three times the square of the area of the side of the hexagon. 2. If the diameter of a circle is extended to C until the extended segment is equal to the radius, then the two tangents from C and their chord of contact form an equilateral triangle. 3. The area of a regular hexagon inscribed in a circle is half the area of an equilateral triangle and three-fourths of the area of a regular hexagon circumscribed to the circle. Proposition 4.16. INSCRIBE A REGULAR FIFTEEN-SIDED POLY- GON IN A GIVEN CIRCLE. It is possible to inscribe a regular, fifteen-sided polygon in a given circle. 4.2. PROPOSITIONS FROM BOOK IV 202 PROOF. We wish to inscribe a regular fifteen-sided polygon in a given circle (oABC). Figure 4.2.19. [4.16] Inscribe a regular pentagon ABCDE in the circle oABC [4.11] and also an equilateral triangle AAGH [4.2]. Join CG. We claim that CG is a side of the required polygon. Since ABCDE is a regular pentagon, the arc ABC is 2\/5 of the circum- ference. Since AAGH is an equilateral triangle, the arc ABG is 1\/3 of the circumference. Hence the arc GC, which is the difference between these two arcs, is equal to \u00a7 \u2014 | = ^ of the entire circumference. Therefore, if chords equal to GC are placed round the circle [4.1], we have a regular fifteen-sided polygon, or quindecagon, inscribed in it. \u25a1 Note: Until the year 1801, no regular polygon could be constructed by lines and circles only except those discussed in this Book IV of Euclid and those obtained from them by the continued bisection of the arcs of which their sides are the chords; but in that year, Gauss proved that if 2n + 1 is a prime number, then regular polygons of 2n + 1 sides are constructible by elementary geometry. Examination question for chapter 4. 1. What is the subject-matter of chapter 4? 2. When is one polygon said to be inscribed in another? 3. When is one polygon said to be circumscribed about another? 4. When is a circle said to be inscribed in a polygon? 5. When is a circle said to be circumscribed about a polygon? 6. What is meant by reciprocal propositions? (Ans. In reciprocal propo- sitions, to every line in one there corresponds a point in the other; and, con- versely, to every point in one there corresponds a line in the other.) 7. Give instances of reciprocal propositions in chapter 4. 4.2. PROPOSITIONS FROM BOOK IV 203 8. What is a regular polygon? 9. What figures can be inscribed in, and circumscribed about, a circle by means of chapter 4? 10. What regular polygons has Gauss proved to be constructible by the line and circle? 11. What is meant by escribed circles? 12. How many circles can be constructed to touch three lines forming a triangle? 13. What is the centroid of a triangle? 14. What is the orthocenter? 15. What is the circumcenter? 16. What is the polar circle? 17. What is the “nine-points circle”? 18. How does a nine-points circle get its name? 19. Name the nine points that a nine-points circle passes through. 20. What three regular figures can be used in filling up the space round a point? (Ans. Equilateral triangles, squares, and hexagons.) 21. If the sides of a triangle are 13, 14, 15 units in length, what are the values of the radii of its inscribed and escribed circles? 22. What is the radius of the circumscribed circle? 23. What is the radius of its nine-points circle? 24. What is the distance between the centers of its inscribed and circum- scribed circles? 25. If r is the radius of a circle, what is the area: (a) of its inscribed equilateral triangle? (b) of its inscribed square? (c) its inscribed pentagon? (d) its inscribed hexagon? (e) its inscribed octagon? (f) its inscribed decagon? 26. With the same hypothesis, find the sides of the same regular figures. Exercises on chapter 4. 1. If a circumscribed polygon is regular, the corresponding inscribed poly- gon is also regular, and conversely. 2. If a circumscribed triangle is isosceles, the corresponding inscribed tri- angle is isosceles, and conversely. 3. If the two isosceles triangles in #2 have equal vertical angles, they are both equilateral. 4.2. PROPOSITIONS FROM BOOK IV 204 4. Divide an angle of an equilateral triangle into five equal parts. 5. Inscribe a circle in a sector of a given circle. 6. Suppose that segments DE, BC of A ABC are parallel: DE \\ BC. Prove that the circles constructed about the triangles A ABC, AADE touch at A. 7. If the diagonals of a cyclic quadrilateral intersect at E, prove that the tangent at E to the circle about the triangle A ABE is parallel to CD. 8. Inscribe a regular octagon in a given square. 9. If a segment of given length slides between two given lines, find the locus of the intersection of perpendiculars from its endpoints to the given lines. 10. If the perpendicular to any side of a triangle at its midpoint meets the internal and external bisectors of the opposite angle at the points D and E, prove that D, E are points on the circumscribed circle. 11. Through a given point P, construct a chord of a circle so that the intercept EE stands opposite a given angle at point X. 12. In a given circle, inscribe a triangle having two sides passing through two given points and the third parallel to a given line. 13. Given four points, no three of which are collinear, construct a circle which is equidistant from them. 14. In a given circle, inscribe a triangle whose three sides pass through three given points. 15. Construct a triangle, being given: (a) the radius of the inscribed circle, the vertical angle, and the perpendic- ular from the vertical angle on the base. (b) the base, the sum or difference of the other sides, and the radius of the inscribed circle, or of one of the escribed circles. (c) the centers of the escribed circles. 16. If F is the midpoint of the base of a triangle, DE the diameter of the circumscribed circle which passes through F, and L the point where a parallel to the base through the vertex meets DE, prove that DL.FE is equal to the square of half the sum and DF.LE is equal to the square of half the difference of the two remaining sides. 17. If from any point within a regular polygon of n sides perpendiculars fall on the sides, their sum is equal to n times the radius of the inscribed circle. 18. The sum of the perpendiculars falling from the angular points of a regular polygon of n sides on any line is equal to n times the perpendicular from the center of the polygon on the same line. 19. If R denotes the radius of the circle circumscribed about a triangle AABC, r, r’ , r” , r'” are the radii of its inscribed and escribed circles; S, 8′ , 8″ are the perpendiculars from its circumcenter on the sides; fi, fi’ ‘, fi” are the 4.2. PROPOSITIONS FROM BOOK IV 205 segments of these perpendiculars between the sides and circumference of the circumscribed circle, then we have the equalities: r \/ + r \/\/ + r \/\/\/ = 4R + r (1) \/x + \/\/ + \/\/’ = 2R-r (2) 8 + 8′ + 8″ = R + r (3) The relation (3) supposes that the circumcenter is inside the triangle. 20. Take a point D from the side BC of a triangle AABC and suppose we construct a transversal EDF through it; suppose we also construct circles about the triangles ADBF, AECD. The locus of their second point of inter- section is a circle. 21. In every quadrilateral circumscribed about a circle, the midpoints of its diagonals and the center of the circle are collinear. 22. Find on a given line a point P, the sum or difference of whose distances from two given points may be given. 23. Find a point such that, if perpendiculars fall from it on four given lines, their feet are collinear. 24. The line joining the orthocenter of a triangle to any point P in the circumference of its circumscribed circle is bisected by the line of co-linearity of perpendiculars from P on the sides of the triangle. 25. The orthocenters of the four triangles formed by any four lines are collinear. 26. If a semicircle and its diameter are touched by any circle either in- ternally or externally, then twice the rectangle contained by the radius of the semicircle and the radius of the tangential circle is equal to the rectangle con- tained by the segments of any secant to the semicircle through the point of intersection of the diameter and touching circle. 27. If p, p’ are radii of two circles touching each other at the center of the inscribed circle of a triangle where each touches the circumscribed circle, prove that 1 1 _ 2 p p’ r and state and prove corresponding theorems for the escribed circles. 28. If from any point in the circumference of the circle, circumscribed about a regular polygon of n sides, segments are constructed to its angular points, the sum of their squares is equal to 2n times the square of the radius. 29. In the same case as the above, if the lines are constructed from any point in the circumference of the inscribed circle, prove that the sum of their 4.2. PROPOSITIONS FROM BOOK IV 206 squares is equal to n times the sum of the squares of the radii of the inscribed and the circumscribed circles. 30. State the corresponding theorem for the sum of the squares of the lines constructed from any point in the circumference of any concentric circle. 31. If from any point in the circumference of any concentric circle perpen- diculars are let fall on all the sides of any regular polygon, the sum of their squares is constant. 32. See #31. For the inscribed circle, the constant is equal to 3n\/2 times the square of the radius. 33. See #31. For the circumscribed circle, the constant is equal to n times the square of the radius of the inscribed circle, together with n times the square of the radius of the circumscribed circle. 34. If the circumference of a circle whose radius is R is divided into sev- enteen equal parts and AO is the diameter constructed from one of the points of division (A), and if p l9 p 2 , ps denote the chords from O to the points of division, A, A 2 , A 8 on one side of AO, then P1P2P4P8 = P3P5P6P7 = R 4 Let the supplemental chords corresponding to pi, p 2 , etc., be denoted by 7*1, r 2 , etc. Then by [3.35, #2], we have that Pin = Rr 2 P2T2 = Rr A p A r A = Rr 8 P8T8 = Rn Hence, pip 2 p4Ps = R 4 – And it may be similarly proved that P1P2P3P4P5P6P7P8 = R 8 Therefore, p 3 p5pep7 = R 4 . 35. If from the midpoint of the segment joining any two of four concyclic points a perpendicular falls on the line joining the remaining two, the six per- pendiculars thus obtained are concurrent. 36. The greater the number of sides of a regular polygon circumscribed about a given circle, the less will be its perimeter. 37. The area of any regular polygon of more than four sides circumscribed about a circle is less than the square of the diameter. 38. Four concyclic points taken three by three determine four triangles, the centers of whose nine-points circles are concyclic. 4.2. PROPOSITIONS FROM BOOK IV 207 39. If two sides of a triangle are given in position, and if their included angle is equal to an angle of an equilateral triangle, the locus of the center of its nine-points circle is a straight line. 40. If in the hypothesis and notation of #34, a, \/3 denote any two suffixes whose sum is less than 8 where a > f3, then PaP(3 = R(p a -P + Pa+fi) For example, pip 4 = R(ps + ps) [3.35, #7]. In the same case, if the suffixes be greater than 8, then Pa ‘ PP = R{pa-(3 pYl-oc-^) For instance, p 8 \u2022 p2 = R(P6 \u2014 Pi) [3.35, #6]. 41. Two lines are given in position. Construct a transversal through a given point, forming with the given lines a triangle of given perimeter. 42. Given the vertical angle and perimeter of a triangle, construct it with any of the following data: (a) The bisector of the vertical angle; (b) the perpendicular from the vertical angle on the base; (c) the radius of the inscribed circle. 43. In a given circle, inscribe a triangle so that two sides may pass through two given points and that the third side may be a maximum or a minimum. 44. If s is the semi-perimeter of a triangle, and if r’, r” ‘, r'” are the radii of its escribed circles, then r > . r ” + r” \u2022 r'” + r'” \u2022 r’ = s 2 45. The feet of the perpendiculars from the endpoints of the base on either bisector of the vertical angle, the midpoint of the base, and the foot of the perpendicular from the vertical angle on the base are concyclic. 46. Given the base of a triangle and the vertical angle, find the locus of the center of the circle passing through the centers of the escribed circles. 47. The perpendiculars from the centers of the escribed circles of a triangle on the corresponding sides are concurrent. 48. If AB is the diameter of a circle, PQ is any chord cutting AB at O, and if the segments AP, AQ intersect the perpendicular to AB at O (at D and E respectively), then the points A, B, D, E are concyclic. 49. If the sides of a triangle are in arithmetical progression, and if R, r are the radii of the circumscribed and inscribed circles, then 6Rr is equal to the rectangle contained by the greatest and least sides. 4.2. PROPOSITIONS FROM BOOK IV 208 50. Inscribe in a given circle a triangle having its three sides parallel to three given lines. 51. If the sides AB, BC, etc., of a regular pentagon is bisected at the points A’, B’, C, D’, E and if the two pairs of alternate sides BC, AE and AB, DE meet at the points A” ‘, E” , respectively, prove that AA”AE” – AA’AE’ = pentagon A’B’C’D’E’ 52. In a circle, prove that an equilateral inscribed polygon is regular; also prove that if the number of its sides are odd, then it is an equilateral circum- scribed polygon. 53. Prove that an equiangular circumscribed polygon is regular; also prove that if the number of its sides are odd, then it is an equilateral inscribed poly- gon. 54. The sum of the perpendiculars constructed to the sides of an equian- gular polygon from any point inside the figure is constant. 55. Express the sides of a triangle in terms of the radii of its escribed circles. CHAPTER 5 Theory of Proportions Chapter 5, like Chapter 2, proves a number of propositions which demon- strate elementary algebraic statements that are more familiar to us in the form of equations. Algebra as we know it had not been developed when Euclid wrote “The Elements”. As such, the results are more of historical importance than practical use (except when they are used in subsequent proofs). As such, Book V appears here in truncated form. 5.1. Definitions 1. A lesser magnitude is said to be a part or submultiple of a greater mag- nitude when the lesser magnitude is contained an exact number of times in the greater magnitude. 2. A greater magnitude is said to be a multiple of a lesser magnitude when the greater magnitude contains the lesser magnitude an exact number of times. 3. A ratio is the mutual relation of two magnitudes of the same kind with respect to quantity. 4. Magnitudes are said to have a ratio to one another when the lesser magnitude can be multiplied so as to exceed the greater. These definitions require explanation, especially [Def. 5.3], which has the fault of conveying no precise meaning \u2014 being, in fact, unintelligible. We recon- sider the above using algebraic terminology: (a) If an integer is divided into any number of equal parts, then one part or the sum of any number of these parts is called a fraction. a c d e a \u2022 * \u25a0> p * Figure 5.1.1. Def. 5.1 209 5.1. DEFINITIONS 210 If the segment AB represents the integer, and if it is divided into four equal parts at the points C, D, E, then AC = 1\/4, AD = 2\/4, and AE = 3\/4 of the whole segment, AB. Thus, a fraction is denoted by two numbers denoted above and below by a horizontal segment; the lower, the denominator, denotes the number of equal parts into which the integer is divided, and the upper, the numerator, denotes the number of these equal parts which are taken. It follows that if the numerator is less than the denominator, the fraction is less than 1. If the numerator is equal to the denominator, the fraction equals to 1. And if the numerator is greater than the denominator, the fraction is greater than 1. It is evident that a fraction is an abstract quantity; that is, that its value is independent of the nature of the integer which is being divided. (b) If we divide each of the equal parts AC, CD, DE, EB into two equal parts, the whole, AB, will be divided into eight equal parts, and we see that AC = 2\/8, AD = 4\/8, AE = 6\/8, and AB = 8\/8. As we saw above, AE = 3\/4 of the integer AB, and we have just demon- strated that AE = 6\/8. Hence, 3\/4 = 6\/8, which we can also obtain by multi- plying the fraction 3\/4 by 2\/2, or 3 _ 3 3 2 _ 6 4 ~ 4 ‘ ~ 4 ‘ 2 ~ 8 Hence we infer generally that multiplying each of the terms of any fraction by 2 does not alter its value. Similarly, it may be shown that multiplying each of the terms of a fraction by any nonzero integer does not alter its value. It follows conversely that dividing each of the terms of a fraction by a nonzero integer does not alter the value. Hence we have the following important and fundamental theorem: “The terms of a fraction can be either both multiplied or both divided by any nonzero integer and in either case the value of the new fraction is equal to the value of the original fraction.” (c) If we take any number, such as 3, and multiply it by any nonzero integer, the product is called a multiple of 3. Thus 6,9,12,15,… are mul- tiples of 3; but 10,13,17,… are not, because the multiplication of 3 by any nonzero integer will not produce them. Conversely, 3 is a submultiple or part of 6, 9, 12, 15,… because it is contained in each of these without a remainder; but not of 10, 13, 17, … because in each case it leaves a remainder. (d) If we consider two magnitudes of the same kind, such as two segments AB, CD in Fig. 5.1.2, and if we suppose that AB = CD, it is evident that if AB is divided into three equal parts and CD is divided into four equal parts 5.1. DEFINITIONS 211 that each of the parts into which AB is divided is equal in length to each of the parts into which CD is divided. D -* Figure 5.1.2. As there are three parts in AB and four in CD, we express this relation by saying that AB has to CD the ratio of 3 to 4, and we denote it as 3 : 4. Hence the ratio 3 : 4 expresses the same idea as the fraction 3\/4. In fact, both are different ways of denoting the same thing: when written 3 : 4, it is called a ratio, and when written as 3\/4, it is fraction. Similarly, it can be shown that every ratio whose terms are commensurabl^] can be converted into a fraction; and, conversely, every fraction can be turned into a ratio. From this explanation, we see that the ratio of any two commensurable magnitudes is the same as the ratio of the numerical quantities which denote these magnitudes. Thus, the ratio of two commensurable lines is the ratio of the numbers which express their lengths measured with the same unit. And this may be extended to the case where the lines are incommensurable. If a is the side of a square and b is its diagonal, the ratio of a : b is a _ 1 __ y\/2 b ~ 7! ” ~2~ When two quantities are incommensurable, such as the diagonal and the side of a square, although their ratio is not equal to that of any two commen- surable numbers, a series of pairs of fractions can be found whose difference is continually diminishing and which ultimately becomes infinitely small. We say that these fractions converge to an irrational numbei^] such as l\/y\/2. In this example, we have that: A 14 141 1414 14142 , 1 { 2 ‘ 20′ 200’ ^00 ‘ ^000′ – } t0 7! In decimal form, this gives us: {0, 0.7, 0.707, 0.7071, 0.707108, …} converges to V2 ]http:\/\/en wikipedia org\/wiki\/Commensurability_ (mathematics) 1http:\/\/en wikipedia org\/wiki\/Irrational_number 5.1. DEFINITIONS 212 By this method and in either form, we can approximate as nearly as we wish to the exact value of the ratio. It is evident we may continue either se- quence as far as we please. For example, if we consider the first member of the fractional sequence as ^ , then we may write the second member as 1Q 1 r g+ fe where k can always be determined. Furthermore, in the case of two incommensurable quantities, two fractions ^ and can always be found where n can be made as large as we wish where one fraction is greater than the irrational fraction and the other fraction is less. To see this, let a and b be the incommensurable quantities. Then, if n ^ m, we cannot have that na = mb. Since any multiple of a must lie between two consecutive multiples of b, such as mb and (m + 1)6, we have that ^ > 1 and 7 ^ttt – because a > c and b L PROPOSITION 5.7. Equal magnitudes have to the same the same ratio; and the same has to equal magnitudes the same ratio. COROLLARY. 1. If any magnitudes are proportional, then they are also proportional inversely. COROLLARY. 2. [5.7] is equivalent to: if a = b, then a = b = kc, and c = qa = qb where q = h 5.2. PROPOSITIONS FROM BOOK V 216 PROPOSITION 5.8. Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater. COROLLARY. 1. [5.8] is equivalent to: if AB > C and D > 0, then AB = C + k where k > 0, and ^ = = % + \u00a3 > %. It follows that \u00a7 > j^, since all quantities are positive. PROPOSITION 5.9. Magnitudes which have the same ratio to the same equal one another; and magnitudes to which the same has the same ratio are equal. COROLLARY. 1. [5.9] is equivalent to: if A = kC and B = kC, then A = B. PROPOSITION 5.10. Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less. COROLLARY. 1. [5.10] is equivalent to: if % and C > 0, then A> B. PROPOSITION 5.11. Ratios which are the same with the same ratio are also the same with one another. COROLLARY. 1. [5.11] is equivalent to: if ^ = % and % = f , then ^ = f . This is the transitive property for fractions. PROPOSITION 5.12. If any number of magnitudes are proportional, then one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. COROLLARY. 1. [5.12] is equivalent to: if A = kB, C = kD, E = kF, then A + C + E = kB + kD + kF = k(B + D + F). 5.2. PROPOSITIONS FROM BOOK V 217 PROPOSITION 5.13. If a first magnitude has to a second the same ratio as a third to a fourth, and the third has to the fourth a greater ratio than a fifth has to a sixth, then the first also has to the second a greater ratio than the fifth to the sixth. COROLLARY. 1. [5.13] is equivalent to: if A = kB, C = kD, C = ID, E = jF and I > j, then k = \/ and so k > j. PROPOSITION 5.14. If a first magnitude has to a second the same ratio as a third has to a fourth, and the first is greater than the third, then the second is also greater than the fourth; if equal, equal; and if less, less. COROLLARY. 1. [5.14] is equivalent to: if A = kB, C = kD, A > C, and k > 0, then kB = A > C = kD and so B > D. If A < C and B C. We wish to show that D > F. Suppose that A = c + m, m > 0. Then A = fcZC, D = klF, and so \u00a3 = ^. Now \u00a3 > 1 since A > C. If D = F, \u00a3 = 1; and if D < F, \u00a3 F. The remaining cases follow mutandis mutatis. PROPOSITION 5.21. If there are three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them is perturbed, then, if the first magnitude is greater than the third, then the fourth is also greater than the sixth; if equal, equal; and if less, less. COROLLARY. 1. The result of [5.21] is the same as the result [5.20]. PROPOSITION 5.22. If there are any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, then they are also in the same ratio. 5.2. PROPOSITIONS FROM BOOK V 219 COROLLARY. 1. [5.22] is equivalent to: if A = kB, B = IC, D = kE, and E = IF, then A = klC and D = klF. PROPOSITION 5.23. If there are three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, then they are also in the same ratio. COROLLARY. 1. The result of [5.23] is the same as the result of [5.22]. PROPOSITION 5.24. If a first magnitude has to a second the same ratio as a third has to a fourth, and also a fifth has to the second the same ratio as a sixth to the fourth, then the sum of the first and fifth has to the second the same ratio as the sum of the third and sixth has to the fourth. COROLLARY. 1. [5.24] is equivalent to: if x = km, u = kn, y = Im, and v = In, then x + y = km + lm = (k + l)m and u + v = kn + In = (k + l)n. PROPOSITION 5.25. If four magnitudes are proportional, then the sum of the greatest and the least is greater than the sum of the remaining two. COROLLARY. 1. [5.25] is equivalent to: let x + y = k(u + v), k > 1, and x = ku. Since x + y = ku + kv, y = kv; and since k > 1, y > v. Examination questions for chapter 5. 1. What is the subject-matter of this chapter? 2. When is one magnitude said to be a multiple of another? 3. What is a submultiple or measure? 4. What are equimultiples? 5. What is the ratio of two commensurable magnitudes? . What is meant by the ratio of incommensurable magnitudes? 7. Give an illustration of the ratio of incommensurables. 8. What are the terms of a ratio called? 9. What is duplicate ratio? 10. Define triplicate ratio. 11. What is proportion? (Ans. Equality of ratios.) 12. How many ratios in a proportion? 5.2. PROPOSITIONS FROM BOOK V 220 13. When is a segment divided harmonically? 14. What are reciprocal ratios? Chapter 5 exercises. 1. A ratio of greater inequality is increased by diminishing its terms by the same quantity, and diminished by increasing its terms by the same quantity. 2. A ratio of lesser inequality is diminished by diminishing its terms by the same quantity, and increased by increasing its terms by the same quantity. 3. If four magnitudes are proportionals, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. 4. If two sets of four magnitudes are proportionals, and if we multiply corresponding terms together, the products are proportionals. 5. If two sets of four magnitudes are proportionals, and if we divide corre- sponding terms, the quotients are proportionals. 6. If four magnitudes are proportionals, their squares, cubes, etc., are pro- portionals. 7. If two proportions have three terms of one respectively equal to three corresponding terms of the other, the remaining term of the first is equal to the remaining term of the second. 8. If three magnitudes are continual proportionals, the first is to the third as the square of the difference between the first and second is to the square of the difference between the second and third. 9. If a line AB, cut harmonically at C and D, is bisected at O, prove that OC, OB, OD are continual proportionals. 10. In the same case, if O’ is the midpoint of CD, prove that OO’ 2 = OB 2 + OD 2 . 11. Continuing from #10, show that AB(AC + AD) = 2AC.AD, or ^ + 1 _ _2_ AD ~ AB 12. Continuing from #10, show that CD(AD + BD) = 2AD.BD, or ^ + 1 _ _2_ AD ~ AC 13. Continuing from #10, show that AB.CD = 2AD.CB. CHAPTER 6 Applications of Proportions When comparing the proportions of areas of triangles, we will use the fol- lowing abbreviation: if we wish to state that the area of AABC divided by the area of ADEF is equal to the area of AGHI divided by the area of AJKL, we will write AABC _ AGHI ADEF ~ AJKL or AABC : ADEF :: AGHI : AJKL This is comparable for how we also write AGHI = AJKL to denote that the area of AGHI is equal to the area of AJKL. 6.1. Definitions 1. Similar polygons are those whose angles are respectively equal and whose sides about the equal angles are proportional. Similar figures agree in shape; if they also agree in size, then they are congruent. If polygons ABC and DEF are similar, we will denote this as ABC ~ DEF. (a) When the shape of a figure is given, it is said to be given in species. Thus a triangle whose angles are given is given in species. Hence, similar figures are of the same species. (b) When the size of a figure is given, it is said to be given in magnitude, such as a square whose side is of given length. (c) When the place which a figure occupies is known, it is said to be given in position. 2. A segment is said to be cut at a point in extreme and mean ratio when the whole segment is to the greater segment as the greater segment is to the lesser segment. 3. If three quantities of the same kind are in continued proportion, the middle term is called a mean proportional between the other two. Magnitudes in continued proportion are also said to be in geometrical progression. 4. If four quantities of the same kind are in continued proportion, the two middle terms are called two mean proportionals between the other two. 221 6.2. PROPOSITIONS FROM BOOK VI 222 5. The altitude of any figure is the length of the perpendicular from its highest point to its base. 6. Two corresponding angles of two figures have the sides about them reciprocally proportional when a side of the first is to a side of the second as the remaining side of the second is to the remaining side of the first. This is equivalent to saying that a side of the first is to a side of the second in the reciprocal ratio of the remaining side of the first to the remaining side of the second. 7. Similar figures are said to be similarly constructed upon given segments when these lines are homologou^] sides of the figures. 8. Homologous points in the planes of two similar figures are such that segments constructed from them to the angular points of the two figures are proportional to the homologous sides of the two figures. See Fig. 6.1.1. 9. The point O in Fig. 6.1.1 is called the center of similitude of the figures. It is also called their double point. Figure 6.1.1. [Def 6.9] See also [6.20, #2] 6.2. Propositions from Book VI PROPOSITION 6.1. PROPORTIONAL TRIANGLES AND PARALLELO- GRAMS. The areas of triangles and parallelograms which have the same al- titude are proportional to their bases. ^Def: “Having the same or a similar relation; corresponding, as in relative position or structure.” Dictionary.com, “homologous,” in Dictionary.com Unabridged. Source location: Ran- dom House, Inc. http:\/\/dictionary.reference.com\/browse\/homologous. Available: http:\/\/dictionary.reference.com. Accessed: May 09, 2013. 6.2. PROPOSITIONS FROM BOOK VI 223 PROOF. The areas of triangles (A ABC, AACD) and of parallelograms (BEBCA, BACDF) which share the same altitude are proportional to their bases (BC, CD). \u25a1 Figure 6.2.1. [6.1] Construct AACB, BEBCA on the base BC, and construct AACD, BACDF on the base CD such that each quadrilateral has the same altitude. Extend BD in both directions to the points H and L. Construct any number of segments BG and GH which are equal in length to the base CB and any number of segments DK and KL which are equal in length to the base CD. Join AG, AH, AK, and AL [1.3]. Since CB = BG = GH, we obtain that AACB = AABG = AAGH [1.38]. Therefore, if CH = k-CB (where k > 1), we also have that AACH = k \u25a0 AACB. Similarly, if CL = m \u2022 CD (where m > 1), we also have that AACL = m \u2022 AACD. Finally, we also have that CH = n \u2022 CL (where n > 0) which implies that AACH = n \u2022 AACL. Hence, we obtain CH : CL CH :m-CD k-CB :m-CD CB : CD AACH : AACL AACH : m \u25a0 AACD k \u25a0 AACB : m \u25a0 AACD AACB : AACD Next, since BEBCA = 2 \u2022 AACB and BACDF = 2 \u2022 AACD [1.41], we have that 2 \u2022 CB : 2 \u2022 CD CB : CD BEBCA : BACDF UEBCA : UACDF \u25a1 PROPOSITION 6.2. PROPORTIONALITY OF SIDES OF TRIANGLES. If a segment is parallel to a side of a triangle, it divides the remaining sides pro- portionally ( when measured from the opposite angle). Conversely, if two sides 6.2. PROPOSITIONS FROM BOOK VI 224 of a triangle, measured from an angle, are cut proportionally, the line joining the points of section is parallel to the third side. PROOF. If a segment (DE) is parallel to a side (BC) of a triangle (AABC), we claim that it divides the remaining sides proportionally (measured from the opposite angle, ZDAE). Conversely, if two sides of a triangle, measured from an angle are cut proportionally, we claim that the segment joining the points of the section is parallel to the third side. 1. Suppose that DE \\ BC in AABC. We wish to show that AD : DB :: AE : EC. Join BE, CD. The triangles ABDE, ACED are on the same base DE and between the same parallels BC, DE. By [1.37], they are equal in area, and, regarding the proportionality of the areas of the triangles, we have that AADE : ABDE :: AADE : ACDE [5.7]. We also have that AADE : ABDE :: AD : DB and AADE : ACDE :: AE : EC, both by [6.1]. It follows that AD : DB :: AE : EC. 2. Now suppose AD : DB :: AE : EC. We wish to show that DE || BC. Let the same construction be made as in part 1. Then we have that AD : BD :: AADE : ABDE and AE : EC :: AADE : ACDE, both by [6.1]. By hypothesis, we also have that AD : DB :: Ai? : \u00a3?C. Hence it follows that AADE : ABDE :: AADE : ACL>\u00a3;. By [5.9], ABDE = ACDE. These triangles also stand on the same base DE as well as on the same side of DE. By [1.39], they stand between the same parallels, and so DE || BC. \u25a1 Figure 6.2.2. [6.2] 6.2. PROPOSITIONS FROM BOOK VI 225 Note: The segment DE may cut the sides AB, AC extended through points B or C or through the angle at A, but it is clear that a separate figure for each of these cases is unnecessary. Exercise. 1. If two segments are cut by three or more parallels, the intercepts on one are proportional to the corresponding intercepts on the other. Proposition 6.3. ANGLES AND PROPORTIONALITY OF TRIANGLES. If a line bisects any angle of a triangle, it divides the opposite side into segments proportional to the adjacent sides. Conversely, if the segments into which a line constructed from any angle of a triangle divides the opposite side is proportional to the adjacent sides, that line bisects the angle. PROOF. If a line (AD) bisects any angle (ABAC) of a triangle (AABC), it divides the opposite side (BC) into segments (BD, DC) proportional to the adjacent sides (BA, AC). Conversely, if the segments (BD, DC) into which a line (AD) constructed from any angle (ABAC) of a triangle divides the opposite side is proportional to the adjacent sides (BA, AC) that line bisects the angle (ABAC). Figure 6.2.3. [6.3] We prove each claim separately: 1. Suppose that AD bisects ABAC of a triangle AABC. Through C, con- struct segment CE || AD to meet BA when BA is extended to the point E. Because BA meets the parallels AD, EC, we have that ABAD = AAEC [1.29]. Also because AC meets the parallels AD, EC, we have that AD AC = AACE. 6.2. PROPOSITIONS FROM BOOK VI 226 By hypothesis, we also have that ZBAD = ZD AC. Therefore, Z ACE = ZAEC, and so AE = AC [1.6]. Again, because AD \\ EC, where EC is one of the sides of the triangle ABEC, we have that BD : DC :: BA: AE [6.2]. Since AE = AC by the above, it follows that BD : DC :: BA: AC. 2. Now suppose that BD : DC BA: AC. We wish to prove that ZBAC is bisected. Let the same construction be made as in part 1. Because AD \\ EC, BA : AE :: BD : DC [6.2]. But BD : DC :: BA : AC by hypothesis. By [5.11], it follows that BA : AE :: BA : AC, and hence AE = AC [5.9]. Therefore, ZAEC = ZACE; we also have that ZACE = ZBAD [1.29] and that ZACE = ZD AC. Hence ZBAD = ZD AC, and so the line AD bisects the angle ZBAC. \u25a1 COROLLARY. 1. [6.3] holds when the line AD is replaced by a ray or seg- ment of appropriate length, mutatis mutandis. Exercises. 1. If the segment AD bisects the external vertical angle ZCAE, then BA : AC :: BD : DC, and conversely. Figure 6.2.4. [6.3], #1 Cut off AE = AC. Join ED. Then the triangles AACD, AAED are evi- dently congruent; therefore the angle ZEDB is bisected, and hence BA : AE :: BD : DE and BA: AC :: BD : DC [6.3]. 6.2. PROPOSITIONS FROM BOOK VI 227 2. #1 has been proved by quoting [6.3]. Prove it independently, and prove [6.3] as an inference from it. 3. The internal and the external bisectors of the vertical angle of a triangle divide the base harmonically. 4. Any segment intersecting the legs of any angle is cut harmonically by the internal and external bisectors of the angle. 5. Any segment intersecting the legs of a right angle is cut harmonically by any two lines through its vertex which make equal angles with either of its sides. 6. If the base of a triangle is given in magnitude and position and if the ratio of the sides is also given, then the locus of the vertex is a circle which divides the base harmonically in the ratio of the sides. 7. If a, b, c denote the sides of a triangle AABC, and D, D’ are the points where the internal and external bisectors of A meet BC, then prove that DD’ = 2abc b 2 -c 2 * 8. In the same case as #7, if E, E’, F, F’ are points similarly determined on the sides CA, AB, respectively, prove that DD’ ^ EE’ ^ FF’ u ^2 7 2 2 DD’ ^ EE’ ^ FF’ u PROPOSITION 6.4. EQUIANGULAR TRIANGLES I. The sides about the equal angles of equiangular triangles are proportional, and those which stand opposite to the equal angles are homologous. PROOF. The sides about the equal angles of equiangular triangles ( ABAC, ACDE) are proportional, and those which are opposite to the equal angles are homologous. Figure 6.2.5. [6.4] 6.2. PROPOSITIONS FROM BOOK VI 228 Let the sides BC, CE which stand opposite to the equal angles ZBAC and ZCDE be constructed so as to form one continuous segment and where the triangles stand on the same side of the segment such that the equal angles ZBCA, ZCED do not share a common vertex. The sum ZABC + ZBCA is less than two right angles, but ZBCA = ZBED by hypothesis. Therefore the sum ZABE + ZBED is less than two right angles, and so the segments AB, ED will meet if extended. Let them meet at F. Again, because ZBCA = ZBEF, we have that CA \\ EE [1.28]. Similarly, BE || CD. Therefore, the figure [HACDF is a parallelogram, and so AC = DF and CD = AF. Because AC \\ FE, BA : AF :: BC : CE [6.2]. But AF = CD, therefore BA : CD :: BC : CE, and so by [5.16], we obtain that BA : BC :: CD : CE. Again, because CD \\ BE, we obtain that BC : CE :: ED : DE. But ED = AC, and so BC : CE :: AC : DE. And by [5.16], BC : AC :: C\u00a3 : DE. Since we have that BA : BC CD : CE and 5C : CA :: C\u00a3 : L>\u00a3, it follows that BA : CA :: CD : and so the sides about the equal angles are proportional. \u25a1 This proposition may also be proved very simply by superposition by Fig. 6.2.2. Construct the two triangles be AABC, AADE and let the second trian- gle AADE be constructed to be placed on AABC so that its two sides AD, AE fall on the sides AB, AC. Since ZADE = ZABC, DE || BC. Hence by [6.2], AD : DB :: AE : EC, and so we have AD AB v. AE AC and AD : AE :: AB : AC by [5.16]. Therefore, the sides about the equal angles ZBAC, ZDAE are proportional, and an analogous result follows for the oth- ers. It can be shown by this proposition that two lines which meet at infinity are parallel. Let \/ denote the point at infinity through which the two given lines pass, and construct any two parallels intersecting them in the points A, B and A! , B’ . Then the triangles AAIB, AA’IB’ are equiangular. Therefore, AI : AB :: A’ I : A’ B’ where the first term of the proportion is equal to the third. By [5.14], the second term AB is equal to the fourth A’B’, and, being parallel to it, the lines AA! , BB’ are parallel [1.43]. Exercises. 1. If two circles intercept equal chords AB, A’B’ on any secant, the tan- gents AT, A’T to the circles at the points of intersection are to one another as the radii of the circles. 6.2. PROPOSITIONS FROM BOOK VI 229 2. If two circles intercept on any secant chords that have a given ratio, the tangents to the circles at the points of intersection have a given ratio, namely, the ratio compounded of the direct ratio of the radii and the inverse ratio of the chords. 3. Being given a circle and a line, prove that a point may be found such that the rectangle of the perpendiculars falling on the line from the points of intersection of the circle with any chord through the point shall be given. 4. If AB is the diameter of a semicircle ADB and CD _L AB, construct through A a chord AF of the semicircle meeting CD at E such that the ratio CE : EE may be given. PROPOSITION 6.5. EQUIANGULAR TRIANGLES II. If two triangles have proportional sides, they are equiangular, and the angles which are equal stand opposite the homologous sides. PROOF. If two triangles (AABC, ADEF) have their sides proportional (BA : AC :: ED : DF; AC : CB :: DF : FE), then they are equiangular and the equal angles stand opposite the homologous sides. : Figure 6.2.6. [6.5] At the points D, E construct the angles ZEDG, ZDEG equal to the angles ABAC, AABC of the triangle AABC. By [1.32], the triangles AABC, AD EG are equiangular. Therefore BA : AC :: ED : DG by [1.4] and BA : AC :: ED : DF by hypothesis. It follows that DG = DF. Similarly, we can show that EG = EF. Hence the triangles AEDF, AEDG have the sides ED, DF in one equal to the sides ED, DG in the other and the 6.2. PROPOSITIONS FROM BOOK VI 230 base EF equal to the base EG. By [1.8], they are equiangular. But the triangle ADEG is equiangular to A ABC. Therefore the triangle ADEF is equiangular to AABC. \u25a1 Observation: In [Def. 6.1], two conditions are laid down as necessary for the similitude of polygons: (a) The equality of angles; (b) The proportionality of sides. Now by [6.4] and [6.5], we see that if two triangles possess either condi- tion, they also possess the other. Triangles are unique in this respect. In all other polygons, one of these conditions may exist without the other. Thus two quadrilaterals may have their sides proportional without having equal angles, or vice verse. PROPOSITION 6.6. If two triangles have one angle in one triangle equal to one angle in the other triangle and the sides about these angles are proportional, then the triangles are equiangular and have those angles equal which stand opposite to the homologous sides. PROOF. If two triangles (AABC, ADEF) have one angle (ZBAC) in one equal to one angle (ZEDF) in the other, and the sides about these angles pro- portional (B A : AC :: ED : DF), then the triangles are equiangular and have those angles equal which stand opposite to the homologous sides. Figure 6.2.7. [6.6] 6.2. PROPOSITIONS FROM BOOK VI 231 We recreate the construction as in [6.5]. Therefore B A : AC :: ED : DG by [6.4] , BA : AC :: ED : DF by hypothesis, and DG = DF. Because ZEDG = ZBAC by construction and ZBAC = ZEDF by hypothe- sis, we have that ZEDG = ZEDF. Given that = with DF in common, the triangles AEDG and AEDF are equiangular. But AEDG is equiangular to ADAC, and so AEDF is equiangular to ABAC. \u25a1 Note: as in the case of [6.4], an immediate proof of this proposition can also be obtained from [6.2]. COROLLARY. 1. If the ratio of two sides of a triangle are given as well as the angle between them, then the triangle is given in species. PROPOSITION 6.7. EQUIANGULAR TRIANGLES III. If two triangles each have one angle equal to one angle in the other, if the sides about two other angles are proportional, and if the remaining angles of the same species (i. e. either both acute or both not acute), then the triangles are similar. PROOF. If two triangles (AABC, ADEF) each have one angle equal to one angle {ABAC = AEDF) in the other, the sides about two other angles (B, E) are proportional (AB : BC :: DE : EF), and the remaining angles (ABC A, ZEFD) of the same species (i. e. either both acute or both not acute), then the triangles are similar (AABC ~ ADEF). A Figure 6.2.8. [6.7] If the angles ZABC and ZDEF are not equal, one must be greater than the other. Wlog, suppose ZABC is the greater and that ZABG = ZDEF. Then the triangles AABG, ADEF have two angles in one equal to two angles in the other and so are equiangular [1.32]. Therefore, AB : BG :: DE : EF [6.4] and AB : BC v. DE : EF by hypothesis. It follows that BG = BC. Hence ZBCG, ZBGC must each be acute [1.17] and ZAGB must be obtuse. It follows 6.2. PROPOSITIONS FROM BOOK VI 232 that ZDFE = ZAGB is obtuse. Since we have shown that ZACB is acute, the angles ZACB, ZDFE are of different species; but by hypothesis, they are of the same species, a contradiction. Hence the angles ZCBA and ZFED are not unequal; that is, ZCBA = ZFED. Therefore, AABG, ADEF are equiangular, and so AABG ~ ADEF. \u25a1 COROLLARY. 1. If two triangles A ABC, ADEF have two sides in one pro- portional to two sides in the other, AB : BC :: DE : EF, and the angles at points A, D opposite one pair of homologous sides are equal, the angles at points C, F opposite the other are either equal or supplemental. This proposition is nearly identical with [6.7]. COROLLARY. 2. If either of the angles at points C, F are right, the other angle must be right. PROPOSITION 6.8. SIMILARITY OF RIGHT TRIANGLES. The triangles formed when a right triangle is divided by the perpendicular from the right angle to the hypotenuse are similar to the whole and to one another. PROOF. The triangles (AACD, ABCD) formed when a right triangle (AACB) is divided by the perpendicular (CD) from the right angle (ZACB) to the hy- potenuse are similar to the whole and to one another (AACD ~ ABCD, AACD ~ AACB, ABCD ~ AACB). Figure 6.2.9. [6.8] Since the two triangles AACD, AACB have the angle ZBAC common, and since ZADC = ZACB because each are right, the triangles are equiangular [1.32]. By [6.4], AACD ~ AACB. Similarly, it can be shown that ABCD ~ AACB. It follows that AACD ~ ABCD. \u25a1 6.2. PROPOSITIONS FROM BOOK VI 233 COROLLARY. 1. The perpendicular CD is a mean proportional between the segments AD, DB of the hypotenuse. For since the triangles AADC, ACDB are equiangular, we have AD : DC :: DC : DB. Hence DC is a mean proportional between AD, DB (Def. 6.3). COROLLARY. 2. BC is a mean proportional between AB, BD, and AC is a mean proportional between AB, AD. COROLLARY. 3. The segments AD, DB are in the duplicate of AC : CB; in other words, AD : DB :: AC 2 : CB 2 . COROLLARY. 4. BA : AD are in the duplicate ratios of BA : AC, and AB : BD are in the duplicate ratio of AB : BC. PROPOSITION 6.9. From a given segment, we may cut off any required sub- multiple. PROOF. From a given segment (AB), we wish to cut off any required sub- multiple. Figure 6.2.10. [6.9] Suppose we wish to cut off a fourth. Construct the segment AF at any acute angle to AB (where AF is made sufficiently long). From AF, choose any point C and cut off the segments CD, DE, EF where each is equal to AC [1.3]. Join BF and construct CG \\ BF. We claim that AG is a fourth of AB. Since CG || BF where BF is the side of AABF, we have that AC : AF :: AG : AB [6.2]. But AC is a fourth of AF by construction, and so AG is a fourth of AB. Since our choice of a fourth was arbitrary any other required submultiple may similarly be cut off. \u25a1 6.2. PROPOSITIONS FROM BOOK VI Note: [1.10] is a particular case of this proposition. 234 Proposition 6.10. SIMILARLY DIVIDED SEGMENTS. We wish to di- vide a given undivided segment similarly to a given divided segment. PROOF. We wish to divide a given undivided segment (AB) similarly to a given divided segment (CD). Figure 6.2.11. [6.10] Construct AG at any acute angle with AB and cut off the parts AH, HI, IG respectively equal to the parts CE, EF, FD of the given divided segment CD. Join BG and construct HK, IL where each is parallel to BG. We claim that AB is divided similarly to CD. Through H, construct HN || AB, cutting IL at M. Now in the triangle AALI, we have that HK || IL. By [6.2], AK : KL :: AH : HI, and by construc- tion, we have that AK : KL :: CE : EF. In AHNG, we have that MI \\ NG. By [6.2], HM : MN :: HI : IG. How- ever, but by [1.34], HM = KL, MN = LB, and HI = EF, and by construction, IG = FD. Therefore, KL : LB :: EF : FD. Hence the segment AB is divided similarly to the segment CD. \u25a1 Exercises. 1. We wish to divide a given segment AB internally or externally in the ratio of two given lines, FG, HJ. 6.2. PROPOSITIONS FROM BOOK VI 235 Figure 6.2.12. [6.11] Through A and B construct any two parallels AC and BD in opposite di- rections. Cut off AC = FG, BD = HJ, and join CD; we claim that CD divides AB internally at E in the ratio of FG : HJ. 2. In #1, if BD’ is constructed in the same direction with AC, then CD will cut AB externally at E in the ratio of FG : HJ. COROLLARY. 1. The two points E, E’ divide AB harmonically. This problem is manifestly equivalent to the following: given the sum or difference of two lines and their ratio, we wish to find the lines. 3. Any line AE’ through the midpoint B of the base DD’ of a triangle DCD’ is cut harmonically by the sides of the triangle and a parallel to the base through the vertex. 4. Given the sum of the squares on two segments and their ratio, find the segments. 5. Given the difference of the squares on two segments and their ratio, find the segments. 6. Given the base and ratio of the sides of a triangle, construct it when any of the following data is given: (a) the area; (b) the difference on the squares of the sides; (c) the sum of the squares on the sides; (d) the vertical angle; 6.2. PROPOSITIONS FROM BOOK VI (e) the difference of the base angles. 236 Proposition 6.11. PROPORTIONAL SEGMENTS I. We wish to find a third proportional segment to two given segments. PROOF. We wish to find a third proportional segment to two given seg- ments ( JK, LM). J \u00bb\u00ab _L _ H Figure 6.2.13. [6.11] Construct any two segments AC, AE at an arbitrary acute angle. (The proof holds with rays or lines, mutatis mutandis.) Cut off AB = JK, BC = LM, and AD = LM. Join BD and construct CE \\ BD. We claim that DE is the required third proportional segment. In ACAE, BD || CE. Therefore AB : BC :: AD : DE by [6.2]. But AB = JK and BC = LM = AD. Therefore JK : LM :: LM : DE. Hence DE is a third proportional to JK and LM. \u25a1 COROLLARY. 5.11.1 Algebraically, this problem can be written as a b_ b x => x = b – a where a, b are fixed positive real numbers and x is a positive real variable. Notes: (1) Another solution can be inferred from [6.8]. For if AD, DC in that proposition are respectively equal to JK and LM, then DB will be the third proportional. Or again, if in Fig. 6.2.9, if AD = JM and AC = LM, then AB will be the third proportional. Hence, we may infer a method of continuing the proportion to any number of terms. 6.2. PROPOSITIONS FROM BOOK VI 237 Exercises. 1. If AAOQ is a triangle where the side AQ is greater than AO, then if we cut off AB = AO, construct BB \\ AO, cut off BC = BB’ , and so on, the series of segments AB, BC, CD, etc., are in continual proportion. o M B 1 I % r: D E” Figure 6.2.14. [6.2, #1] 2. (AB-BC) : AB :: AB : Att. This is evident by constructing MB’ \\ Ail. PROPOSITION 6.12. PROPORTIONAL SEGMENTS II. We wish to find a fourth proportional to three given segments. PROOF. We wish to find a fourth proportional to three given segments (AK, BM, CP). Construct any two segments DE, DF at an arbitrary acute angle. (The proof holds with rays or lines, mutatis mutandis.) Cut off DG = AK, GE = BM, and DH = CP. Join GH and construct EE || GH [1.31]. We claim that HF is the required fourth proportional segment. Figure 6.2.15. [6.12] 6.2. PROPOSITIONS FROM BOOK VI 238 In ADEF, GH \\ EF. Therefore, DG : GE :: DH : HF by [6.2]. But DG = AK, GE = BM, and DH = CP. Therefore AK : BM :: CP : HF, and so HF is a fourth proportional to AK, BM, and CP. \u25a1 Alternatively: PROOF. Take two segments AD, BC which intersect at O. FIGURE 6.2.16. [6.12], Alternative proof Construct OA = JK, OB = LM, OC = MP and the circle oABC through the points A, B, C [4.5]. Extend AO through to the point D on the circum- ference of oABC. We claim that OD is the fourth proportional required. The demonstration is evident from the similarity of the triangles AAOB and ACOD \u25a1 COROLLARY. 5.12.1 Algebraically, this problem can be written as a c_ b x X = ^ a where a, b, and c are fixed positive real numbers and xisa positive real variable. PROPOSITION 6.13. PROPORTIONAL SEGMENTS III We wish to find a mean proportional between two given segments. 6.2. PROPOSITIONS FROM BOOK VI 239 PROOF. We wish to find a mean proportional between two given segments (EF, GH). J F _j3 H Figure 6.2.17. [6.13] Construct segment AC such that AC = EF + GH and cut parts AB, BC respectively equal to EF, GH. (The proof holds on rays or lines, mutatis mu- tandis.) On AC, construct a semicircle ADC. Construct BD JL AC, meeting the semicircle at D. We claim that BD is the mean proportional required. Join AD, DC. Since ADC is a semicircle, the angle ZADC is right [3.31]. Hence, since AADC is a right triangle and BD is a perpendicular from the right angle on the hypotenuse, BD is a mean proportional between AB, BC [6.8, Cor. 1]. That is, BD is a mean proportional between EF and GH (or EF : BD :: \u00a3L> : GH). \u25a1 COROLLARY, i. Algebraically, we have that a x x b => x 2 = a 6 => x = Vab where a, b are positive fixed real numbers and x is a positive real variable. Exercises. 1. Another solution may be inferred from [6.8, Cor. 2]. 2. If through any point within a circle a chord is constructed which is bisected at that point, its half is a mean proportional between the segments of any other chord passing through the same point. 3. The tangent to a circle from any external point is a mean proportional between the segments of any secant passing through the same point. 6.2. PROPOSITIONS FROM BOOK VI 240 4. If through the midpoint C of any arc of a circle, a secant is constructed cutting the chord of the arc at D and the circle again at E, the chord of half the arc is a mean proportional between CD and CE. 5. If a circle is constructed touching another circle internally and with two parallel chords, the perpendicular from the center of the former on the diameter of the latter, which bisects the chords, is a mean proportional between the two extremes of the three segments into which the diameter is divided by the chords. 6. If a circle is constructed touching a semicircle and its diameter, the diameter of the circle is a harmonic mean between the segments into which the diameter of the semicircle is divided at the point of intersection. 7. State and prove the proposition corresponding to #5 for external contact of the circles. Proposition 6.14. EQUIANGULAR PARALLELOGRAMS. We wish to prove that: 1. Equiangular parallelograms (IHACB, BCGDE) which are equal in area have sides about the equal angles which are reciprocally proportional; that is, AC : CE :: GC : CB. 2. Equiangular parallelograms which have the sides about the equal angles reciprocally proportional are equal in area. PROOF. We claim that: 1. Equiangular parallelograms ([HHACB, [ICGDE) which are equal in area have the sides about the equal angles reciprocally proportional; that is, AC : CE :: GC : CB. 2. Equiangular parallelograms which have the sides about the equal an- gles reciprocally proportional are equal in area. Figure 6.2.18. [6.14] 6.2. PROPOSITIONS FROM BOOK VI 241 We prove each claim separately: 1. Let [HHACB, [HCGDE be so placed as to form one segment AE such that the equal angles ZACB, ZECG stand vertically opposite. Since ZACB = ZECG, add ZBCE to each, and we obtain the sum ZACB + ZBCE = ZECG + ZBCE. But ZACB + Z\u00a3G\u00a3 equals two right angles [1.13], and so ZECG + ZBCE equals two right angles. By [1.14], BC, CG form one segment. We complete the parallelogram BBCEF. Again, since the parallelograms [HHACB, BCGDE are equal in area by hypothesis, we have that BHACB : BBCEF :: BCGDE : BBCEF [5.7] AC:CE :: BHACB:BBCEF [6.1] BCGDE : BBCEF :: GG : C\u00a3 [6.1] Therefore, we have that AC : C\u00a3 :: GC : C5; that is, the sides about the equal angles are reciprocally proportional. 2. Let AC : CE :: GC : C\u00a3. We wish to prove that the parallelograms RHACB, BCGDE are equal in area. Let the same construction be made as in part 1: BHACB : BBCEF [6.1] GC : CB [6.1] GC : CB by hypothesis Therefore, BHACB : BBCEF :: BCGDE : RBCEF, and hence BHACB = BCGDE [5.9]. \u25a1 AC:C\u00a3 BCGDE : BBCEF AC :CE Alternatively: PROOF. Join i\/i}, RD. The area of the parallelogram BHACB = 2 \u2022 AH BE, and the area of the parallelogram BCGDE = 2 \u2022 ABDE. Therefore AHBE = ABDE, and by [1.39.], HD \\ BE. Hence HB : \u00a3F :: DE : \u00a3F; that is, AC : C\u00a3 :: GC : CB. Part 2 may be proved by reversing this demonstration. Another demonstration of this proposition may be obtained by extending the lines HA and DG to meet at \/. Then by [1.43], the points \/, G, F are collinear, and the proposition is evident. \u25a1 Proposition 6.15. EQUAL TRIANGLES. 6.2. PROPOSITIONS FROM BOOK VI 242 1. Two triangles equal in area which have one angle in one triangle equal to one angle in the other triangle have reciprocally proportional sides about these angles. 2. Two triangles which have one angle in one triangle equal to one angle in the other triangle and the sides about these angles reciprocally proportional are equal in area. PROOF. We wish to prove that: 1. Two triangles equal in area {AACB, ADCE) which have one angle {ABC A) in one equal to one angle {ADCE) in the other have reciprocally pro- portional sides about these angles. 2. Two triangles which have one angle in one triangle equal to one angle in the other triangle and the sides about these angles reciprocally proportional are equal in area. Figure 6.2.19. [6.15] We prove each claim separately: 1. Let the equal angles be placed as to be vertically opposite such that AC, CD forms the segment AD; then it may be demonstrated, as in the previous proposition, that BC, CE form one segment. Join BD. Since the triangles AACB = ADCE, we have that AACB : ABCD :: ADCE : ABCD [5.7] AACB : ABCD :: AC : CD [6.1] ADCE : ABCD :: EC:CB [6.1] Therefore, AC : CD :: EC : CB, or the sides about the equal angles are reciprocally proportional. 2. If AC : CD :: EC : CB, we wish to prove that AACB = ADCE. 6.2. PROPOSITIONS FROM BOOK VI 243 Let the same construction be made; then we have that AC: CD :: AACB : ABCD [6.1] EC : CB :: ADCE : ABCD [6.1] AC : CD :: EC :CB (by hypothesis) Therefore, AAC\u00a3 : ABCD :: AL>C\u00a3 : ABCD, and so AACB = ADCE [5.9]. \u25a1 This proposition might have been appended as a corollary to [6.14] since the triangles are the halves of equiangular parallelograms; it may also be proven by joining AE and showing that it is parallel to BD. Proposition 6.16. PROPORTIONAL RECTANGLES. We wish to show that: 1. If four segments are proportional, the rectangle contained by the extremes is equal in area to the rectangle contained by the means. 2. If the rectangle contained by the extremes of four segments is equal in area to the rectangle contained by the means, the four segments are propor- tional. PROOF. We wish to show that: 1. If four segments (AB, CD, LM, NP) are proportional, the rectangle (AB.NP) contained by the extremes is equal in area to the rectangle (CD.LM) contained by the means. 2. If the rectangle contained by the extremes of segments are equal to the rectangle contained by the means, the four lines are proportional. L H H P Figure 6.2.20. [6.16] We solve each claim separately: 1. Suppose that the segments AB, CD, LM, NP are proportional (that is, AB : CD :: LM : NP). Construct AH = NP and CI = LM at right angles 6.2. PROPOSITIONS FROM BOOK VI 244 to AB and CD, respectively, and complete the rectangles. Because AB : CD :: LM : NP by hypothesis, we have that AB : CD :: CI : Aff . Notice that the parallelograms BHABG, BICDK are equiangular and the sides about their equal angles are reciprocally proportional. By [6.14], they are equal in area. Since since AH = NP, we have that BHABG = AB.NP. Similarly, BICDK = CD.LM. It follows that AB.NP = CD.LM, or the rectangle contained by the extremes is equal to the rectangle contained by the means. 2. Suppose that AB.NP = CD.LM. We now wish to prove AB : CD :: LM : NP. We carry out the same construction as in part 1: because AB.NP = CD.LM, AH = NP, and CI = LM, we have that BHABG = BICDK. Since these par- allelograms are equiangular, the sides about their equal angles are reciprocally proportional. Therefore AB : CD :: CI : AH, or AB : CD :: LM : NP. \u25a1 Alternatively: PROOF. Place the four segments in a concurrent position so that the ex- tremes form one continuous segment and the means form a second continuous segment. FIGURE 6.2.21. [6.16], Alternative proof Place the four segments in the order AO, BO, OD, OC. Join AB, CD. Be- cause AO : OB :: OD : OC and ZAOB = ZDOC, the triangles AAOB, ACOD are equiangular. Hence, the four points A, B, C, D are concyclic, and so by [3.35], AO.OC = BO.OD. \u25a1 COROLLARY. 6.16.1 Algebraically, the above result states in part that if a c b = d then ad = be 6.2. PROPOSITIONS FROM BOOK VI 245 Proposition 6.17. LINES AND RECTANGLES. 1. If three segments are proportional, the rectangle contained by the end- points is equal in area to the square of the mean. 2. If the rectangle contained by the endpoints of three segments is equal in area to the square of the mean, then the three segments are proportional. PROOF. We wish to show that: 1. If three segments (AB, CD, GH) are proportional (AB : CD :: CD : GH), then the rectangle (AB.GH) contained by the extremes is equal in area to the square of the mean (CD 2 ). 2. If the rectangle contained by the endpoints of three segments are equal in area to the square of the mean, then the three lines are proportional. 4 iP ^ p c h J- L I Figure 6.2.22. [6.17] We prove each claim separately: 1. Suppose that AB : CD v. CD : GH and that CD = EF. By hypoth- esis, we have that AB : CD :: EF : GH. By [6.16], AB.GH = CD.EF. But CD.EF = CD 2 . Therefore, AB.GH = CD 2 ; that is, the rectangle contained by the extremes is equal to the square of the mean. 2. Now suppose that AB.GH = CD 2 . Under the same construction, since AB.GH = CD.EF, we have that AB : CD :: EF : GH where CD = EF. It follows that AB : CD :: CD : GH; that is, the three segments are proportionals. \u25a1 Note: This proposition may also be inferred as a corollary to [6.16]. Exercises. 1. Construct a Corollary similar to Corollary 6.16.1 which states the re- sults of [6.17] algebraically. 2. If a segment CD bisects a vertical angle at a point C of any triangle AACB, its square added to the rectangle AD.DB contained by the segments of the base is equal to the rectangle contained by the sides. 6.2. PROPOSITIONS FROM BOOK VI 246 Construct a circle about the triangle and extend CD to meet it at E. Clearly, the triangles AACD, AECB are equiangular. By [6.4], AC : CD :: CE : CB, and by [3.35], we obtain AC.CB = CE.CD = CD 2 + CL>.L>\u00a3 = CD 2 + AD.DB 3. If the segment CD bisects the external vertical angle of any triangle AACB, its square subtracted from the rectangle AD.DB is equal to AC.CB. (See Fig. 6.2.23). 4. The rectangle contained by the diameter of the circumscribed circle, and the radius of the inscribed circle of any triangle, is equal to the rectangle contained by the segments of any chord of the circumscribed circle passing through the center of the inscribed. (See Fig. 6.2.23). Let O be the center of the inscribed circle. Join OB, construct the per- pendicular OG, and construct construct the diameter EE of the circumscribed circle. Now we have that ZABE = ZECB and ZABO = ZOBC [3.27]; there- fore ZEBO = sum of ZOCB, ZOBC = ZEOB. Hence, EB = EO. Again, the triangles AEBF, AOGC are equiangular because ZEFB = ZECB and ZEBF = ZOGC (since each are right). Therefore, EE : EB :: OC : OG, from which it follows that EF.OG = EB.OC = EO.OC 5. #3 may be extended to each of the escribed circles of AACB. 6. The rectangle contained by two sides of a triangle is equal to the rec- tangle contained by the perpendicular and the diameter of the circumscribed circle. (See Fig. 6.2.24). 6.2. PROPOSITIONS FROM BOOK VI 247 Figure 6.2.24. [6.17, #5] Let CE be the diameter and join AE. Then the triangles AAGF, ADCB are equiangular; hence AC : CE :: CD : CB, and therefore AC.CB = CD.CE. 7. If a circle passing through the angle at point A of a parallelogram BABCD intersects the two sides AB, AD again at the points F, G and the diagonal AC again at F, then AB.AE + AD. AG = AC.AF. (See Fig. 6.2.25). Figure 6.2.25. [6.17, #6] Join FF, FG, and make the angle ZABH = ZAFE. Then the triangles AABH, AAFE are equiangular. Therefore AB : AH :: AF : AE. Hence, AB.AE = AF.AH. Again, it is clear that the triangles ABCH, AGAF are equiangular, and therefore BC : CH :: AF : AG. Hence 5G.AG = AF.CH, or AD. AG = AF.CH; but we have proved AB.AE = AF.AH. Therefore AD. AG + AB.AE = AF.AC. 8. If \u00a3>F, DF are parallels to the sides of a triangle AA\u00a3?G from any point D in the base, then AB.AE + AG. AF = AD 2 + BD.DC. Hint: deduce this from #6. 9. If through a point O within a triangle AA5G parallels FF, Gi7, IF are constructed to the sides, the sum of the rectangles of their segments is equal to the rectangle contained by the segments of any chord of the circumscribing circle passing through O. 6.2. PROPOSITIONS FROM BOOK VI 248 Notice that AO.AL = AB.AK + AC.AE and AO 2 = AG.AK + AH.AE-GO.OH Hence, AO.OL = BG.AK + CiJ.AE + GO.Otf or AO.OL = \u00a3O.OF + \/O.Oif + GO.Otf 10. The rectangle contained by the side of an inscribed square standing on the base of a triangle and the sum of the base and altitude is equal to twice the area of the triangle. 11 The rectangle contained by the side of an escribed square standing on the base of a triangle and the difference between the base and altitude is equal to twice the area of the triangle. 12. If from any point P in the circumference of a circle a perpendicular is constructed to any chord, its square is equal to the area of the rectangle contained by the perpendiculars from the endpoints of the chord on the tangent at P. 13. If O is the point of intersection of the diagonals of a cyclic quadrilateral ABCD, the areas of the four rectangles AB.BC, BD.CD, CD. DA, DA.AB are proportional to lengths of the four segments BO, CO, DO, AO. 14. Ptolemy’s Theorem: the sum of the rectangles of the opposite sides of a cyclic quadrilateral ABCD is equal in area to the rectangle contained by its diagonals. 6.2. PROPOSITIONS FROM BOOK VI 249 Figure 6.2.27. [6.17, #13] Construct the circle oABC on quadrilateral ABCD. AD AO = ZCAB. Then the triangles AD AO, ACAB are equiangular. Therefore, AD : DO :: AC : CB, and it follows that AD.BC = AC. DO. Again, the triangles AD AC, AOAB are equiangular and CD : AC :: BO : AB. Therefore AC.CD = AC. BO, and so AD.BC + AB.CD = AC.BD. 15. If the quadrilateral ABCD is not cyclic, prove that the three rectangles AB.CD, BC.AD, AC.BD are proportional to the three sides of a triangle which has an angle equal to the sum of a pair of opposite angles of the quadrilateral. 16. Prove by using [6.11] that if perpendiculars fall on the sides and di- agonals of a cyclic quadrilateral, from any point in the circumference of the circumscribed circle, the rectangle contained by the perpendiculars on the di- agonals is equal to the rectangle contained by the perpendiculars on either pair of opposite sides. 17. If AB is the diameter of a semicircle and PA, PB are chords from any point P in the circumference, and if a perpendicular to AB from any point C meets PA, PB at D and E and the semicircle at point F, then CF is a mean proportional between CD and CE. PROPOSITION 6.18. CONSTRUCTION OF A SIMILAR POLYGON We wish to construct a polygon on a given segment which is similar to a given polygon and similarly placed. PROOF. Let polygon CDEFG and segment AB be given. We wish to con- struct a polygon on AB which is similar to polygon CDEFG and which is sim- ilarly placed. 6.2. PROPOSITIONS FROM BOOK VI 250 Figure 6.2.28. [6.18] Join CE, CF, and construct triangle AABH on AB equiangular to ACDE and similarly placed as regards CD; that is, construct ZABH = ZCDE and ZBAH = ZDCE. Also construct the triangle AHAI equiangular to AECF and similarly placed. Finally, construct the triangle AIAJ equiangular and similarly placed with AFCG. We claim that ABHIJ is the required polygon. By construction, it is evident that the figures are equiangular, and it is only required to prove that the sides about the equal angles are proportional. Because the triangle AABH is equiangular to ACDE, we have that AB : BH :: CD : DE [6.4]. Hence the sides about the equal angles at points B and D are proportional. Again from the same triangles, we have BH : HA :: DE : EC, and from the triangles AIHA, AFEC, we have HA : HI :: EC : EF. There- fore, BH : HI :: DE : EF, or the sides about the equal angles ZBHI, ZDEF are proportional. This result follows about the other equal angles, mutatis mutandis. Hence by [Def. 6.1], the figures are similar. \u25a1 Observation: in the foregoing construction, the segment AB is homologous to CD, and it is evident that we may take AB to be homologous to any other side of the given figure CDEFG. Again, in each case, if the figure ABHIJ is turned round the segment AB until it falls on the other side, it will still be similar to the figure CDEFG. Hence on a given line AB, there can be constructed two figures each similar to a given figure CDEFG and having the given segment AB homologous to any given side CD of the given figure. The first of the figures thus constructed is said to be directly similar, and the second is said to be inversely similar to the given figure. 6.2. PROPOSITIONS FROM BOOK VI 251 COROLLARY. 1. Twice as many polygons may be constructed on AB similar to a given polygon CDEFG as that figure has sides. COROLLARY. 2. If the figure ABHIJ is applied to CDEFG so that the point A coincides with C and that the line AB is placed along CD, then the points H, I, J will be respectively on the segments CE, CF, CG. Also, the sides BH, HI, IJ of the one polygon will be respectively parallel to their homologous sides DE, EF, FGofthe other COROLLARY. 3. If segments constructed from any point O in the plane of a figure to all its angular points are divided in the same ratio, the lines joining the points of division will form a new figure similar to and having every side parallel to the homologous side of the original. Note: [6.19] is the first of Euclid’s Proposition in which the technical term “duplicate ratio” occurs. Most students find it difficult to understand either Euclid’s proof or his definition. Due to this, we follow Euclid’s proof with John Casey’s alternative proof which makes use of a new definition of the duplicate ratio of two lines: the ratio of the squares constructed on these segments. PROPOSITION 6.19. RATIOS OF SIMILAR TRIANGLES. Similar trian- gles have their areas to one another in the duplicate ratio of their homologous sides. PROOF. We claim that similar triangles (AABC, ADEF) have their areas to one another in the duplicate ratio of their homologous sides. L A D F Figure 6.2.29. [6.191(a) 6.2. PROPOSITIONS FROM BOOK VI 252 Take BG as a third proportional to BC, EF [6.11]; that is, BC : BG :: \u00a3G : \u00a3F Join AG. Then because AABC ~ ADEF, AB : BC :: DE : EF; hence we also have that AB : DE :: \u00a3?C : EF [5.16]. Since we also have that BC : EF :: EF : \u00a3G, by [5.11] we also have that AB : DE :: EF : BG. Hence the sides of the triangles AABG, ADEF about the equal angles at points B, E are reciprocally proportional, and therefore At4.BG, ADEF are equal in area. Again, since the segments BC, EF, BG are continual proportionals, BC : BG is in the duplicate ratio of BC : EF [Def. 5.10]; but BC : BG :: AABC : AABG. Therefore AABC : AABG in the duplicate ratio of BC : EF. But we have shown that the triangle AABG = ADEF. Therefore, the triangle AABC is to the triangle ADEF in the duplicate ratio of BC : EF. \u25a1 Casey’s proof: PROOF. Suppose that AABC ~ ADEF. On AB and DE, construct squares {[HAGHB and HDLME, respectively), and through points C and F construct segments parallel to AB and DE. Extend AG, BR, DL, and EM to points J, \/, O, and AT, respectively; this constructs the rectangles BJABI and BODEN. Figure 6.2.30. [6.19] (\/?) Clearly, the triangles A J AC, AODF are equiangular. Hence: J A: AC :: OD : DF [6.4] and AC:AB :: DF : DE [6.4] : AB :: : DE Since A\u00a3? = AG and DE = DL by construction, we also have that J A : AG :: : DL. 6.2. PROPOSITIONS FROM BOOK VI 253 Again, J A : AG OD : DL BJABI : BAGHB BJABI : BAGHB [6.1] and BODEN : BDLME [6.1] => BODEN : BDLME Therefore, BJABI : BODEN :: BAGHB : BDLME by [5.16], and hence A ABC : ADEF :: AB 2 : Di? 2 \u25a1 Exercises. 1. If one of two similar triangles has a side that is 50% longer than the homologous sides of the other, what is the ratio of their areas? 2. When the inscribed and circumscribed regular polygons of any common number of sides to a circle have more than four sides, the difference of their areas is less than the square of the side of the inscribed polygon. PROPOSITION 6.20. DIVISION OF SIMILAR POLYGONS. Similar poly- gons may be divided: 1. into the same number of similar triangles; 2. such that the corresponding triangles have the same ratio to one another which the polygons have; 3. such that the polygons are to each other in the duplicate ratio of their homologous sides. PROOF. Let ABHIJ, CDEFG be the polygons, and let the sides AB, CD be homologous. Join AH, AI, CE, CF. We prove each claim separately: Figure 6.2.31. [6.20] 6.2. PROPOSITIONS FROM BOOK VI 254 1. We claim that similar polygons may be divided into the same number of similar triangles. Since the polygons are similar, they are equiangular and have proportional sides about their equal angles [Def. 6.1]. Hence, the angle at B is equal to the angle at D, and AB : BH :: CD : DE. By [6.6], the triangle AABH is equiangular to ACDE, and so ZBHA = ZDEC. But ZBHI = ZDEF by hypothesis; therefore, ZAHI = ZCEF. Again, because the polygons are similar, IH : HB :: EE : ED; and since AABH ~ ACDE, HB : HA :: ED : EC. It follows that IH : HA :: EE : \u00a3C, and we have shown that ZIHA = ZFEC. Therefore, AIHA, AFEC are equiangular. Similarly, it can be shown that the remaining triangles are equiangular. 2. We claim that similar polygons may be divided such that the correspond- ing triangles have the same ratio to one another which the polygons have. Since AABH ~ ACDE, we have that AABH : ACDE in the duplicate ratio of AH : CE [6.19] Similarly, A AH I : ACEF in the duplicate ratio of AH : CE Hence, AABH : ACDE = AAHI : ACEF [5.9]. Similarly, AHI : CEF = AIJ : CFG. In these equal ratios, the triangles AABH, AAHI, AAIJ are the an- tecedents, the triangles ACDE, ACEF, ACFG are the consequents, and any one of these equal ratios is equal to the ratio of the sum of all the antecedents to the sum of all the consequents [5.7]. Therefore, AABH : ACDE :: polygon ABHIJ : polygon CDEFG 3. We claim that similar polygons are to each other in the duplicate ratio of their homologous sides. The triangle AABH : ACDE in the duplicate ratio of AB : CD [6.19] Hence by (2), polygon ABHIJ : polygon CDEFG in the duplicate ratio of AB : CD \u25a1 6.2. PROPOSITIONS FROM BOOK VI 255 COROLLARY. 1. The perimeters of similar polygons are to one another in the ratio of their homologous sides. COROLLARY. 2. As squares are to similar polygons, the duplicate ratio of two segments is equal to the ratio of the squares constructed on them. COROLLARY. 3. Similar portions of similar figures bear the same ratio to each other as the wholes of the figures. COROLLARY. 4. Similar portions of the perimeters of similar figures are to each other in the ratio of the whole perimeters. Exercises. 1. If two figures are similar, to each point in the plane of one there will be a corresponding point in the plane of the other. Figure 6.2.32. [6.20, #1] Let ABCD, A’B’C’D’ be the two figures and P a point inside of ABCD. Join AP, BP, and construct a triangle AA \/ P \/ B \/ on A’B’ similar to AAPB; clearly, segments from P’ to the angular points of A’B’C’D’ are proportional to the lines from P to the angular points of ABCD. 2. If two figures are directly similar and in the same plane, there is in the plane called a homologous point with respect to the other (which may be regarded as belonging to either figure). Let AB, A’B’ be two homologous sides of the figures and C their point of intersection. Through the two triads of points A, A C and B, B f , C construct two circles intersecting again at the point O: we claim that O is the required point. Clearly, AOAB ~ AOAB and either may be turned round the point O, so that the two bases, AB, A’B’ will be parallel. 3. Two regular polygons of n sides each have n centers of similitude. 4. If any number of similar triangles have their corresponding vertices lying on three given lines, they have a common center of similitude. 5. If two figures are directly similar and have a pair of homologous sides parallel, every pair of homologous sides will be parallel. Definition: Figures such as those in #5 are said to be homothetic. 6. If two figures are homothetic, the segments joining corresponding angu- lar points are concurrent, and the point of concurrence is the center of simili- tude of the figures. 7. If two polygons are directly similar, either may be turned round their center of similitude until they become homothetic, and this may be done in two different ways. 8. Two circles are similar figures. 6.2. PROPOSITIONS FROM BOOK VI 257 Figure 6.2.34. [6.20, #8] Let O, O’ be their centers, and let the angle ZAOB be made indefinitely small so that the arc AB may be regarded as a straight line; construct ZAOB = ZA’O’B’. We claim that the triangles AAOB ~ AA’O’B’ Again, construct the angle ZBOC indefinitely small and set ZB’O’C = ZBOC. Hence, ABOC ~ AB’O’C. Proceeding in this way, we see that the circles can be divided into the same number of similar elementary triangles. Hence the circles are similar figures. 9. Sectors of circles having equal central angles are similar figures. 10. As any two points of two circles may be regarded as homologous, two circles have in consequence an infinite number of centers of similitude. Their locus is the circle, whose diameter is the line joining the two points for which the two circles are homothetic. 11. The areas of circles are to one another as the squares of their diam- eters. For they are to one another as the similar elementary triangles into which they are divided, and these are as the squares of the radii. 12. The circumferences of circles are proportional to their diameters (see [6.20, Cor. 1]). 13. The circumference of sectors having equal central angles are propor- tional to their radii. Hence if a, a’ denote the arcs of two sectors which stand opposite equal angles at the centers, and if r, r’ are their radii, then we have that a\/r = a! \/r’. 14. The area of a circle is equal to half the rectangle contained by the circumference and the radius. (This is evident by dividing the circle into ele- mentary triangles, as in #8.) 15. The area of a sector of a circle is equal to half the rectangle contained by the arc of the sector and the radius of the circle. PROPOSITION 6.21. TRANSITIVITY OF SIMILAR POLYGONS. Polygons which are similar to the same figure are similar to one another. 6.2. PROPOSITIONS FROM BOOK VI 258 PROOF. We claim that polygons (ABC, DEF) which are similar to the same figure (GHI) are similar to one another (that is, the property of simi- larity is transitive). FIGURE 6.2.35. [6.21] Note that the polygons need not be triangles. Since ABC ~ GHI, they are equiangular and have the sides about their equal angles proportional. Similarly, DEF and GHI are equiangular and have the sides about their equal angles proportional. Hence ABC and DEF are equiangular and have the sides about their equal angles proportional; or, ABC ~ DEF. \u25a1 COROLLARY. 1. Two similar polygons which are homothetic to a third are homothetic to one another. 1. If three similar polygons are respectively homothetic, then their three centers of similitudes are collinear. PROPOSITION 6.22. PROPORTIONALITY OF FOUR SEGMENTS TO THE POLYGONS CONSTRUCTED UPON THEM. If four segments are proportional, then if any pair of similar polygons are similarly constructed on the first and second segments, and if any other pair of similar polygons are constructed on the third and fourth segments, then these figures are proportional. Conversely, if we have that a polygon constructed on the first of four seg- ments is similar and similarly constructed to the polygon constructed on the second segment as a polygon constructed on the third segment is similar and similarly constructed to the polygon constructed on the fourth segment, then the four lines are proportional. Exercise. 6.2. PROPOSITIONS FROM BOOK VI 259 PROOF. Construct four proportional segments (AB, CD, EF, GH) and sim- ilar polygons (AABK, ACDL) constructed from AB and CD. Also construct similar polygons (SMEFI, BNGHJ) from the third and fourth segments (EI, GJ). Si Figure 6.2.36. [6.22] We prove each claim separately: 1. We claim that AABK : ACDL :: BHEFI : BNGHJ. Suppose that the segments AB, CD, EF, GH are proportional. Then we have that AABK : ACDL BHEFI : BNGHJ AB : CD AB 2 : CD 2 [6.20] EF 2 : GH 2 [6.20] EF : GH (by hypothesis) Since we have that AB 2 : CD 2 :: EF 2 : GH 2 by [5.22, Cor. 1], it follows that AABK : ACDL :: BHEFI : BNGHJ. 2. Now suppose that ABK : CDL :: BHEFI : BNGHJ. We wish to show that AB :CD :: EF : G#. Notice that BHEFI : BNGHJ AB 2 : CD 2 AB 2 : CT> 2 [6.20] \u00a3F 2 : GH 2 [6.20] \u00a3F 2 : G# 2 [5.22, Cor. 1] Hence, AB : CD :: EF : GH. B PROPOSITION 6.23. EQUIANGULAR PARALLELOGRAMS. Equiangular parallelograms are to each other as the rectangles contained by their sides about a pair of equal angles. 6.2. PROPOSITIONS FROM BOOK VI 260 PROOF. Equiangular parallelograms (BHABD, BBGEC) are to each other as the rectangles contained by their sides about a pair of equal angles (or X : Z :: AB.BD : BC.BG). H M m A E Figure 6.2.37. [6.23] Let the two sides AB, BC about the equal angles ZABD, ZCBG be placed so as to form one segment AC; clearly, as in [6.14], GB, BD also form the seg- ment GD. Complete the parallelogram [HDBCF. Denoting the parallelograms BHADB, BDBCF, BBGEC by X, Y, Z, respectively, we have that X : Y :: AB : BC [6.1] Y : Z :: BD : BG [6.1] Hence, XY : YZ :: AB.BD : BC.BG, or X : Z :: AB.BD : BC.BG. \u25a1 Exercises. 1. Triangles which have one angle of one equal or supplemental to one angle of the other are to one another in the ratio of the rectangles of the sides about those angles. 2. Two quadrilaterals whose diagonals intersect at equal angles are to one another in the ratio of the rectangles of the diagonals. PROPOSITION 6.24. SIMILAR PARALLELOGRAMS ABOUT THE DIAG- ONAL. In any parallelogram, every two parallelograms which are about a di- agonal are similar to the whole and to one another. PROOF. In any parallelogram (Hi? ADC), every two parallelograms (EI\u00a3ACF, EIJFHC) which are about a diagonal are similar to the whole and to one an- other. 6.2. PROPOSITIONS FROM BOOK VI 261 Figure 6.2.38. [6.24] Since the parallelograms BBADC, BEAGF have a common angle, they are equiangular [1.34], and all we are required to prove that that the sides about the equal angles are proportional. Since the segments EF, are parallel, the triangles AAEF, AABC are equiangular; by [6.4], AE : EF :: AB : BC, and the other sides of the par- allelograms are equal to AE, EF; AB, BC. Hence the sides about the equal angles are proportional, and therefore BEAGF ~ E]BADC. The parallelo- grams BEAGF, [HJFHC may also be shown to be similar in the same way. \u25a1 COROLLARY. 1. The parallelograms REAGF, UJFRC, BBADC are, re- spectively homothetic. PROPOSITION 6.25. CONSTRUCTION OF A POLYGON EQUAL IN AREA TO A GIVEN FIGURE AND SIMILAR TO A SECOND GIVEN FIGURE. PROOF. We wish to construct a polygon equal to a given polygon, ALMN, and similar to a second polygon, BCD. 6.2. PROPOSITIONS FROM BOOK VI 262 On any side of the polygon BCD (wlog, we choose BC), construct the rec- tangle BBJEC = BCD [1.45], and on CE construct the rectangle BCEKF = ALMN. Between BC, CF, find a mean proportional GH and on it construct the polygon GHI ~ BCD [6.18] so that BC and GH may be homologous sides. We claim that GHI is the required polygon. We have that BC : GF :: GF : CF; therefore BC : CF is in the duplicate ratio of BC : GH [Def 5.10]. Since BCD ~ GHI, BCD : GHI is in the duplicate ratio of BC : GH [6.20]. We also have that BC : CF :: BBJEC : BCEKF. Hence, BBJCE : BCEKF :: BCD : GiLT. But the rectangle BBJEC is equal to the polygon BCD; therefore, BCEKF = GHI. Recall that by construction BCEKF = ALMN. It follows that GHI = ALMN and is similar to BCD. Hence, it is the required polygon. \u25a1 Alternatively: PROOF. Construct the squares BEFJK, BLMNO equal in area to the polygons BCD and ALMN, respectively [2.14]. Find GH, a fourth proportional to EF, LM, and BC [6.12]. 6.2. PROPOSITIONS FROM BOOK VI 263 FIGURE 6.2.40. [6.25], alternate proof. Note that GHI need not be a triangle. On GH, construct the polygon GHI similar to the polygon BCD [6.18] such that BC and GH are homologous sides. We claim that GHI is the required polygon. Because EF : LM :: BC : GH by construction, we have that EFJK : LMNO :: BCD : GHI [6.22]. But EFJK = BCD by construction; therefore, LMNO = GHI. But we also have that LMNO = APQS by construction. Therefore GHI = APQS and is similar to BCD. \u25a1 PROPOSITION 6.26. PARALLELOGRAMS ON A COMMON ANGLE. If two similar and similarly situated parallelograms have a common angle, they stand on the same diagonal. PROOF. If two similar and similarly situated parallelograms ([HAEFG, BABCD) have a common angle (ZGAF), we claim that they stand on the same diagonal. Figure 6.2.41. [6.26] 6.2. PROPOSITIONS FROM BOOK VI 264 Construct the diagonals AF, AC. Because s BAEFG ~ BABCD, they can be divided into the same number of similar triangles [6.20]. Hence, AGAF ~ ACAD, and it follows that ZGAF = ZCAD. Hence, AC must pass through the point F and hence the parallelograms are about the same diagonal. \u25a1 Observation: [6.26] is the converse of [6.24] and may have been misplaced in an early edition of Euclid. The following would be a simpler statement of result: “If two homothetic parallelograms have a common angle, they are about the same diagonal.” Proposition 6.27. INSCRIBING A PARALLELOGRAM IN A TRIAN- GLE I. PROOF. Construct AABC. Bisect the side AC at point P opposite to the angle at point B. Through P, construct PE, PF parallel to the remaining sides of the triangle AABC. We claim that BEBFP is the required parallelogram. Figure 6.2.42. [6.27] Take any other point D on AC, construct DG, DH parallel to the sides of AABC, and construct CK \\ AB. Extend EP, GD to meet CK at K and J, and extend HD to meet PK at \/. Since AC is bisected at P, EK is also bisected in P. By [1.36], BEGOP = BPOJK. Therefore, BEGOP > BIDJK; but BIDJK = BOFHD [1.43], and so BEGOP > BOFHD. To each add BGBFO, and we have that BEBFP > BGBHD. Hence, BEBFP is the maximum parallelogram which can be in- scribed in the triangle AABC. B 6.2. PROPOSITIONS FROM BOOK VI 265 COROLLARY. 1. The maximum parallelogram exceeds any other parallelo- gram about the same angle in the triangle by the area of the similar parallel- ogram whose diagonal is the line between the midpoint P of the opposite side and the point D, which is the corner of the other inscribed parallelogram. COROLLARY. 2. The parallelograms inscribed in a triangle and having one angle common with it are proportional to the rectangles contained by the segments of the sides of the triangle made by the opposite corners of the paral- lelograms. COROLLARY. 3. The parallelogram in [6.27] has proportions AC : GH :: AC 2 : AD. DC. Proposition 6.28. INSCRIBING A PARALLELOGRAM IN A TRIAN- GLE II. We wish to inscribe in a given triangle a parallelogram equal to a given polygon not greater than the maximum inscribed parallelogram and having an angle common with the triangle. PROOF. We wish to inscribe a parallelogram in a given triangle (AABC) equal in area to a given polygon (XYZ) not greater than the maximum in- scribed parallelogram and having an angle (at B) in common with the triangle. FIGURE 6.2.43. [6.28] Note that XYZ need not be a triangle. Bisect the side AC at P, opposite to point B. Construct PF, PE parallel to the sides AB, BC. Then [IEBFP is the maximum parallelogram that can be inscribed in the triangle ABC [6.27]. If XYZ = REBFP, the problem is solved. 6.2. PROPOSITIONS FROM BOOK VI 266 Otherwise, extend EP and construct C J parallel to PF; then construct the parallelogram BKLMN equal to the difference between the areas between the polygons BPJCF and XYZ [6.25] and similar to BPJCF such that the sides PJ and KL will be homologous. Cut off PI = KL. Construct IH \\ AB, cutting AC at D, and construct DG parallel to BC. We claim that BGBHD is the required parallelogram. Since the parallelograms BPFCJ, BPODI stand on the same diagonal, they are similar [6.24]; however, BPFCJ ~ BKLMN by construction, and therefore BPODI ~ BKLMN. Also by construction, their homologous sides, PI and KL, are equal. Hence by [6.20], BPODI = BKLMN. Now, BPODI is the difference between BEBFP and BGBHD [6.27, Cor. 1], and BKLMN is the difference between BPJCF and XYZ by construction. Therefore, the dif- ference between EIPJCF and XYZ is equal to the difference between BEBFP and BGBHD. However, BEBFP = BPJCF. Hence, BGBHD = XYZ. B Proposition 6.29. ESCRIBING A PARALLELOGRAM TO A TRIAN- GLE. We wish to escribe to a given triangle a parallelogram equal to a given polygon and having an angle common with an external angle of the triangle. PROOF. We wish to escribe to a given triangle (AABC) a parallelogram equal in area to a given polygon (XYZ) and having an angle common with an external angle (at B) of the triangle. L \u25a0 Figure 6.2.44. [6.29] The construction is the same as [6.28] except that we construct the par- allelogram BKLMN so that its area is equal to the sum of parallelograms 6.2. PROPOSITIONS FROM BOOK VI 267 BPODI and BXYZ. Make PI = KL and construct IH \\ AB; the remaining construction takes place as in [6.28]. Now as in [2.6], it can be shown that the parallelogram BBGDH is equal in area to the gnomon OHJ that is, the area of BBGDH equals to the difference between the parallelograms BPODI and BPFCJ. By construction, this is also the difference between BKLNM and BPFCJ, which is equal in area of XYZ. Notice that BBGDH is escribed to the triangle AABC and has an angle common with the external angle at B. Hence, the proof. \u25a1 PROPOSITION 6.30. We wish to divide a given segment into its “extreme and mean ratio” PROOF. We wish to divide a given segment (AB) into its “extreme and mean ratio.” AC B Figure 6.2.45. [6.30] Divide AB at C so that the rectangle AB.BC = AC 2 [2.11]. We claim that C is the required point. Because the rectangle AB.BC = AC 2 , we have that AB : AC :: AC : BC [6.17]. Hence AB is cut in extreme and mean ratio at C [Def. 6.2]. \u25a1 Exercises. 1. If the three sides of a right triangle are in continued proportion, the hypotenuse is divided in extreme and mean ratio by the perpendicular from the right angle on the hypotenuse. 2. In the same case as #1, the greater segment of the hypotenuse is equal to the least side of the triangle. 6.2. PROPOSITIONS FROM BOOK VI 268 3. The square on the diameter of the circle constructed about the triangle formed by the points F, H, D (see Fig. 6.2.46) is equal to six times the square on the segment FD. Proposition 6.31. AREA OF SQUARES ON A RIGHT TRIANGLE. If any similar polygon is similarly constructed on the three sides of a right trian- gle, the polygon on the hypotenuse is equal in area to the sum of the areas of those polygons constructed on the two other sides. PROOF. If any similar polygon is similarly constructed on the three sides of a right triangle (AABC), we claim that the polygon on the hypotenuse is equal in area to the sum of the areas of those polygons constructed on the two other sides. y K c Cvl Figure 6.2.47. [6.31] Construct CD _L AB [1.12]. Because AABC is a right triangle and CD is constructed from the right angle perpendicular to the hypotenuse, BD : AD is in the duplicate ratio of BA : AC [6.8, Cor. 4]. Again, because the polygons constructed on BA, AC are similar, they are in the duplicate ratio of BA : AD [6.20]. Hence by [5.11], BA : AD :: figure constructed on BA : figure constructed on AC Similarly, AB : BD :: figure constructed on AB : figure constructed on BC By [5.24], AB : AD + BD :: figure constructed on AB : sum of the figures on AC, BC 6.2. PROPOSITIONS FROM BOOK VI 269 But AB = AD+BD. Therefore the polygon constructed on the segment AB is equal in area to the sum of the similar polygons constructed on the segments AC and BC. \u25a1 Alternatively: PROOF. Denote the sides by a, b, c, and the polygons by a, \/3, 7. Because the polygons are similar, by [6.20] we have that a : 7 :: a 2 : c 2 . Therefore, 2 a _ a z 7 c 2 Similarly, P _ 7 c 2 from which it follows that a + f3 _ a 2 + b 2 7 c 2 But a 2 + b 2 = c 2 by [1.47]. Therefore, a + \/? = 7; or, the sum of the polygons on the sides is equal to the polygon on the hypotenuse. \u25a1 Exercise. 1. If semicircles are constructed on supplemental chords of a semicircle, the sum of the areas of the two crescents thus formed is equal to the area of the triangle whose sides are the supplemental chords and the diameter. PROPOSITION 6.32. FORMATION OF TRIANGLES. If two triangles which have two sides of one triangle proportional to two sides of the other triangle and the contained equal angles are joined at an angle so as to have their homologous sides parallel, the remaining sides are in the same segment. PROOF. If two triangles (AABC, ACDE) which have two sides of one tri- angle proportional to two sides of the other triangle (AB : BC :: CD : DE) and the contained equal angles (at points B, D) are joined at an angle (at C) so as to have their homologous sides parallel, the remaining sides (AC, CE) are in the same segment (AE). 6.2. PROPOSITIONS FROM BOOK VI 270 Figure 6.2.48. [6.32] Because the triangles AABC, ACDE have equal angles at B and D and the sides about these angles are proportional, (AB : BC :: CD : DE), they are equiangular [6.6]. Therefore, ABAC = ZDCE. To each add ZACD, and we have ABAC + ZACD = ZDCE + ZACD. But ZBAC + ZACD is two right angles [1.29]. It follows that ZDCE + ZACD is two right angles. Hence by [1.47], AC, CE are in the same segment. \u25a1 PROPOSITION 6.33. RATIOS OF EQUAL TRIANGLES. In equal circles, angles at the centers or at the circumferences have the same ratio to one another as the arcs on which they stand. This also holds true for sectors. PROOF. In equal circles (oABC, oDEF), angles at the centers (ZBOC, ZEPF) or at the circumferences (ZBAC, ZEDF) have the same ratio to one another as the arcs {BC, EF) on which they stand. This also holds true for sectors (BOC, EPF). Figure 6.2.49. [6.33] We prove each claim separately: 1. We claim that BC : EF :: ZBAC : ZEDF. 6.2. PROPOSITIONS FROM BOOK VI 271 Take any number of arcs CG, GH in the first circle, each equal in length to BC. Join OG, OH. In the second circle, take any number of arcs FI, I J, each equal to EF. Join IP, JP. Because the arcs BC = CG = GH, we have that ZBOC = ZCOG = ZGOH [3.27]. Therefore, the arc BH and ZBOH are equimultiples of the arc BC and ZBOC. Similarly, it may be shown that the arc E J and ZEPJ are equimultiples of the arc EF and ZEPF. Again, since the circles are equal, it is evident that ZBOH is greater than, equal to, or less than ZEPJ according to whether the arc BH is greater than, equal to, or less than the arc E J. We have four mag- nitudes: the arc BC, the arc EF, the angle ZBOC, and the angle ZEPF. We have also taken equimultiples of the first and third (the arc BH and ZBOH) and other equimultiples of the second and fourth (the arc E J and ZEPJ) and we have proved that, according as the multiple of the first is greater than, equal to, or less than the multiple of the second, the multiple of the third is greater than, equal to, or less than the multiple of the fourth. Hence by [Def. 5.5], BC : EF :: ZBOC : ZEPF. Again, since the angle ZBAC is half the angle ZBOC [3.20.] and ZEDF is half the angle ZEPF, we have that ZBOC : ZEPF :: ZBAC : ZEDF [5.15] from which it follows that BC : EF :: ZBAC : ZEDF [5.11]. 2. We claim that sector BOC : sector EPF :: BC : EF. We make the same construction as in part 1. Since the arc BC is equal in length to CG, ZBOC = ZCOG. Hence the sectors BOC, COG are con- gruent (see Observation, [3.29]); therefore, they are equal in length. Simi- larly, the sectors COG, GOH are equal in length. Hence there are as many equal sectors as there are equal arcs; therefore, the arc BH and the sector BOH are equimultiples of the arc BC and the sector BOC. Similarly, it may be shown that the arc EJ and the sector EPJ are equimultiples of the arc EF and the sector EPF. And it is evident by superposition that if the arc BH is greater than, equal to, or less than the arc EJ, the sector BOH is greater than, equal to, or less than the sector EPJ. Hence by [Def. 5.5], arc BC : arc EF :: sector BOC : sector EPF. \u25a1 Alternative proof to part 2): PROOF. Sector BOC = (| rectangle by arc BC) and the radius of the circle oABC [6.20, #14], and sector EPF = (2- rectangle contained by the arc EF and the radius of the circle oEDF). Since the circles are equal, their radii are equal. Hence, sector BOC : sector EPF :: arc BC : arc EF. \u25a1 6.2. PROPOSITIONS FROM BOOK VI 272 Examination questions for chapter 6. 1. What is the subject-matter of chapter 6? (Ans. Application of the theory of proportion.) 2. What are similar polygons? 3. What do similar polygons agree in? 4. How many conditions are necessary to define similar triangles? 5. How many conditions are necessary to define similar polygons of more than three sides? 6. When is a polygon said to be given in species? 7. When in magnitude? 8. When in position? 9. What is a mean proportional between two lines? 10. Define two mean proportionals. 11. What is the altitude of a polygon? 12. If two triangles have equal altitudes, how do their areas vary? 13. How do these areas vary if they have equal bases but unequal alti- tudes? 14. If both bases and altitudes differ, how do the areas vary? 15. When are two lines divided proportionally? 16. If in two segments which are divided proportionally a pair of homol- ogous points coincide with their point of intersection, what property holds for the lines joining the other pairs of homologous points? 17. Define reciprocal proportion. 18. If two triangles have equal areas, prove that their perpendiculars are reciprocally proportional to the bases. 19. What is meant by inversely similar polygons? 20. If two polygons are inversely similar, how can they be changed into polygons which are directly similar? 21. Give an example of two triangles inversely similar. (Ans. If two lines passing through any point O outside a circle intersect it in pairs of points A, A’; B, B’, respectively, the triangles AOAB, AOA’B’ are inversely similar.) 22. What point is it round which a polygon can be turned so as to bring its sides into positions of parallelism with the sides of a similar polygon? (Ans. The center of similitude of the two polygons.) 23. How many polygons similar to a given polygon of sides can be con- structed on a given line? 24. How many centers of similitude can two regular polygons of n sides each have? (Ans. n centers, which lie on a circle.) 25. What are homothetic polygons? 6.2. PROPOSITIONS FROM BOOK VI 273 26. How do the areas of similar polygons vary? 27. What proposition is [6.19] a special case of? 28. Define Philo’s line. 29. How many centers of similitude do two circles have? Exercises for chapter 6. 1. If in a fixed triangle, we construct a variable side parallel to the base, the locus of the points of intersection of the diagonals of the trapezium that is cut off from the triangle is the median that bisects the base. 2. Find the locus of the point which divides in a given ratio the several lines constructed from a given point to the circumference of a given circle. 3. Two segments AB, XY, are given in position: AB is divided at C in the ratio m : n and parallels AA’, BB’ , CC are constructed in any direction meeting XY in the points A’, B’, C . Prove that (m + n)CC’ = nAA’ + mBB’ 4. Three concurrent lines from the vertices of a triangle AABC meet the opposite sides in A’, B’, C . Prove AB’.BC’.CA’ = A’B.B’C.C’A. 5. If a transversal meets the sides of a triangle AABC at the points A’, B’ , C, prove AB’.BC’.CA’ = -A’B.B’C.C’A. 6. If on a variable segment AC which is constructed from a fixed point A to any point B on the circumference of a given circle, a point C is taken such that the rectangle AB.AC is constant, prove that the locus of C is a circle. 7. If D is the midpoint of the base BC of a triangle AABC, E the foot of the perpendicular, L is the point where the bisector of the angle at A meets BC, and H the point of intersection of the inscribed circle with BC, prove that DE.HL = HE.HD. 8. As in #7, if K is the point of intersection with BC of the escribed circle, which touches the other extended sides, prove that LH.BK = BD.LE. 9. If R, r, r’, r”, r'” are the radii of the circumscribed, the inscribed, and the escribed circles of a plane triangle, d, d! , d”, d!” the distances of the center of the circumscribed circle from the centers of the others, then R 2 = d 2 + 2Rr = d’ 2 -2Rr’, etc. 10. As in #9, prove that 12R 2 = d 2 + d’ 2 + d” 2 + d'” 2 . 11. If p’, p”, p'” denote the altitudes of a triangle, then: (1) – 1 + ^ + = 1 p’ p” p'” r (2) + 4tt – 1 = A (etc.) v \/ p\/\/ i p\/\/\/ p r’ v 7 (3) 2 = I–L (etc.) (4) | = ^ + ^ (etc.) 6.2. PROPOSITIONS FROM BOOK VI 274 12. In a given triangle, inscribe another of given form which has one of its angles at a given point in one of the sides of the original triangle. 13. If a triangle of given form moves so that its three sides pass through three fixed points, the locus of any point in its plane is a circle. 14. Suppose that the angle at point A and the area of a triangle AABC are given in magnitude. If the point A is fixed in position and the point B move along a fixed line or circle, then the locus of the point C is a circle. 15. One of the vertices of a triangle of given form remains fixed, and the locus of another is a segment or circle. Find the locus of the third. 16. Find the area of a triangle: (a) in terms of its medians; (b) in terms of its perpendiculars. 17. If two circles touch externally, their common tangent is a mean propor- tional between their diameters. 18. If there are three given parallel lines and two fixed points A, B, and if the lines connecting A and B to any variable point in one of the parallels intersects the other parallels at the points C and D, E and F, respectively, prove that CF and DE each pass through a fixed point. 19. If a system of circles pass through two fixed points, any two secants passing through one of the points are cut proportionally by the circles. 20. Find a point O in the plane of a triangle AABC such that the diameters of the three circles about the triangles AOAB, AOBC, AOCA may be in the ratios of three given segments. 21. Suppose that ABCD is a cyclic quadrilateral, and the segments AB, AD, and the point C are given in position. Find the locus of the point which divides BD in a given ratio. 22. If CA, CB are two tangents to a circle and BE _L AD (where AD is the the diameter through A), then prove that CD bisects BE. 23. If three segments from the vertices of a triangle AABC to any interior point O meet the opposite sides in the points A’, B’, C, prove that OA’ OB’ OC _ ~AA’ + ~BB’ + ~CC’ ~ 1 24. If three concurrent lines OA, OB, OC are cut by two transversals in the two systems of points A, B, C; A’, B O, respectively, then prove that AB OC BC OA _ CA OB A’B’ ‘ ~OC’ B’C ‘ ~OA ~C r A ‘ OB* 25. The line joining the midpoints of the diagonals of a quadrilateral cir- cumscribed to a circle: 6.2. PROPOSITIONS FROM BOOK VI 275 (a) divides each pair of opposite sides into inversely proportional segments; (b) is divided by each pair of opposite segments into segments which when measured from the center are proportional to the sides; (c) is divided by both pairs of opposite sides into segments which when measured from either diagonal have the same ratio to each other. 26. If CD, CD’ are the internal and external bisectors of the angle at C of the triangle AACB, the three rectangles AD.DB, AC.CB, AD.BD are proportional to the squares of AD, AC, AD and are: (a) in arithmetical progression, if the difference of the base angles is equal to a right angle; (b) in geometrical progression if one base angle is right; (c) in harmonic progression if the sum of the base angles is equal to a right angle. 27. If a variable circle touches two fixed circles, the chord of contact passes through a fixed point on the line connecting the centers of the fixed circles. Figure 6.2.50. Ch. 6, #27 Let 0,0′ be the centers of the two fixed circles; O the center of the variable circle; A, B the points of contact. Let AB and 00′ meet at C, and cut the fixed circles again in the points A’, B’ respectively. Join A’O, AO, BO’. Then AO, BO’ meet at O” [3.11]. Now because the triangles OAA’, O” AB are isosceles, the angles 0″BA = O” AB = OA’ A. Hence OA’ \\ O’B; therefore OC : O’C :: OA’ : O’ B is in a given ratio. Hence, C is a given point. 28. If DD’ is the common tangent to the two circles, then DD’ 2 = AB’.A’B. 29. If R denotes the radius of O” and p, p’ the radii of O, O’, then DD’ 2 : AB 2 :: (R\u00b1 p)(R\u00b1 p’) : R 2 where the choice of sign depends on the nature of the contacts. (This result follows from #28.) 30. If four circles are tangential to a fifth, and if we denote by 12 the common tangent to the first and second, etc., then 12.34 + 23.14 = 13.24. 6.2. PROPOSITIONS FROM BOOK VI 276 31. The inscribed and escribed circles of any triangle are all touched by its nine-points circle. 32. The four triangles which are determined by four points, taken three by three, are such that their nine-points circles have one common point. 33. If a, b, c, d denote the four sides, and D, D’ the diagonals of a quadrilat- eral, prove that the sides of the triangle, formed by joining the feet of the per- pendiculars from any of its angular points on the sides of the triangle formed by the three remaining points, are proportional to the three rectangles ac, bd, DD f . 34. Prove the converse of Ptolemy’s theorem (see [6.17], #13). 35. Construct a circle which: (a) passes through a given point, and touch two given circles; (b) touches three given circles. 36. If a variable circle touches two fixed circles, the tangent to it from their center of similitude through which the chord of contact passes is of constant length (see Fig. 6.2.50). 37. If the lines AD, BD’ are extended (see Fig. 6.2.50), they meet at a point on the circumference of O”, and the line 0″P is perpendicular to DD’ . 38. If A, B are two fixed points on two lines given in position, and A f , B’ are two variable points such that the ratio AA! : BB’ is constant, the locus of the point dividing A’B’ in a given ratio is a segment. 39. If a segment EF divides proportionally two opposite sides of a quadri- lateral, and a segment GH the other sides, each of these is divided by the other in the same ratio as the sides which determine them. 40. In a given circle, inscribe a triangle such that the triangle whose an- gular points are the feet of the perpendiculars from the endpoints of the base on the bisector of the vertical angle and the foot of the perpendicular from the vertical angle on the base may be a maximum. 41. In a circle, the point of intersection of the diagonals of any inscribed quadrilateral coincides with the point of intersection of the diagonals of the circumscribed quadrilateral whose sides touch the circle at the angular points of the inscribed quadrilateral. 42. Through two given points describe a circle whose common chord with another given circle may be parallel to a given line, or pass through a given point. 43. Being given the center of a circle, describe it so as to cut the legs of a given angle along a chord parallel to a given line. 6.2. PROPOSITIONS FROM BOOK VI 277 44. If concurrent lines constructed from the angles of a polygon of an odd number of sides divide the opposite sides each into two segments, the product of one set of alternate segments is equal in area to the product of the other set. 45. If a triangle is constructed about a circle, the lines from the points of contact of its sides with the circle to the opposite angular points are concurrent. 46. If a triangle is inscribed in a circle, the tangents to the circle at its three angular points meet the three opposite sides at three collinear points. 47. The external bisectors of the angles of a triangle meet the opposite sides in three collinear points. 48. Construct a circle touching a given line at a given point and cutting a given circle at a given angle. Definition: the center of mean position of any number of points A, B, C, D, etc., is a point which may be found as follows: bisect the line joining any two points A, B at G. Join G to a third point C; divide GC at H so that GH = GC. Join H to a fourth point D and divide HD at K, so that HK = HD, and so on. The last point found will be the center of mean position of the given points. 49. The center of mean position of the angular points of a regular polygon is the center of figure of the polygon. 50. The sum of the perpendiculars let fall from any system of points A, B, C, D, etc., whose number is n on any line L, is equal to n times the perpendic- ular from the center of mean position on L. 51. The sum of the squares of segments constructed from any system of points A, B, C, D, etc., to any point P exceeds the sum of the squares of seg- ments from the same points to their center of mean position, O, by nOP 2 . 52. If a point is taken within a triangle so as to be the center of mean position of the feet of the perpendiculars constructed from it to the sides of the triangle, the sum of the squares of the perpendiculars is a minimum. 53. Construct a quadrilateral being given two opposite angles, the diago- nals, and the angle between the diagonals. 54. A circle rolls inside another of double its diameter; find the locus of a fixed point in its circumference. 55. Two points, C, D in the circumference of a given circle are on the same side of a given diameter. Find a point P in the circumference at the other side of the given diameter, AB, such that PC, PD may cut AB at equal distances from the center. 56. If the sides of any polygon be cut by a transversal, the product of one set of alternate segments is equal to the product of the remaining set. 6.2. PROPOSITIONS FROM BOOK VI 278 57. A transversal being constructed cutting the sides of a triangle, the lines from the angles of the triangle to the midpoints of the segments of the transversal intercepted by those angles meet the opposite sides in collinear points. 58. If segments are constructed from any point P to the angles of a trian- gle, the perpendiculars at P to these segments meet the opposite sides of the triangle at three collinear points. 59. Divide a given semicircle into two parts by a perpendicular to the diameter so that the radii of the circles inscribed ob them may have a given ratio. 60. From a point within a triangle, suppose that perpendiculars fall on the sides; find the locus of the point when the sum of the squares of the lines joining the feet of the perpendiculars is given. 61. If a circle makes given intercepts on two fixed lines, the rectangle contained by the perpendiculars from its center on the bisectors of the angle formed by the lines is given. 62. If the base and the difference of the base angles of a triangle are given, the rectangle contained by the perpendiculars from the vertex on two lines through the midpoint of the base, parallel to the internal and external bisectors of the vertical angle, is constant. 63. The rectangle contained by the perpendiculars from the endpoints of the base of a triangle on the internal bisector of the vertical angle is equal to the rectangle contained by the external bisector and the perpendicular from the middle of the base on the internal bisector. 64. State and prove the corresponding theorem for perpendiculars on the external bisector. 65. Suppose that R, R! denote the radii of the circles inscribed in the trian- gles into which a right triangle is divided by the perpendicular from the right angle on the hypotenuse. If c is the hypotenuse and s is the semi-perimeter, R 2 + R’ 2 = (s-c) 2 . 66. If A, B, C, D are four collinear points, find a point O in the same line with them such that OA.OD = OB.OC. 67. Suppose the four sides of a cyclic quadrilateral are given; construct it. 68. Being given two circles, find the locus of a point such that tangents from it to the circles may have a given ratio. 69. If four points A, B, C, D are collinear, find the locus of the point P at which AB and CD stand opposite equal angles. 6.2. PROPOSITIONS FROM BOOK VI 279 70. If a circle touches internally two sides of a triangle, CA, CB, and its circumscribed circle, the distance from C to the point of intersection on either side is a fourth proportional to the semi-perimeter, CA, and CB. 71. State and prove the corresponding theorem for a circle touching the circumscribed circle externally and two extended sides. 72. Pascal’s Theorem: if the opposite sides of an irregular hexagon ABCDEF inscribed in a circle are extended until they meet, the three points of intersec- tion G, H, I are collinear. See Fig. 6.2.51. FIGURE 6.2.51. Ch. 6, #72. Pascal’s Theorem Join AD. Construct a circle about the triangle AADI, cutting the extended segments AF, CD, if necessary, at K and L. Join IK, KL, LI. By [3.21], we have that ZKLG = ZFCG = AG AD. Therefore KL \\ CF. Similarly, LI \\ CH and KI || FH; hence the triangles AKLI, AFCH are homothetic, and so the lines joining corresponding vertices are concurrent. Therefore, the points \/, H, G are collinear. 73. If two sides of a triangle circumscribed to a given circle are given in position with the third side variable, the circle constructed about the triangle touches a fixed circle. 6.2. PROPOSITIONS FROM BOOK VI 280 74. If two sides of a triangle are given in position, and if the area is given in magnitude, two points can be found at each of which the base stands opposite a constant angle. 75. If a, b, c, d denote the sides of a cyclic quadrilateral and s its semi- perimeter, prove that its area = yj (s-a) (s-b) (s-c) (s-d) . 76. If three concurrent lines from the angles of a triangle AABC meet the opposite side in the points A f , B’, C, and the points A f , B’, C are joined, forming a second triangle AA’B’C, then AABC : AA’B’C :: AB.BC.CA : 2AB’.BC’.CA’ 77. In the same case as #76, find the diameter of the circle circumscribed about the triangle AABC = AB’.BC’.CA’ divided by the area of A’ B’C. 78. If a quadrilateral is inscribed in one circle and circumscribed to an- other, the square of its area is equal to the product of its four sides. 79. If on the sides AB, AC of a triangle AABC we take two points D, E on their connecting segment such that BD _ AE _ DE AD _ ~CE ~ lEF then prove that the triangle BFC = 2ADE. 80. If through the midpoints of each of the two diagonals of a quadrilateral we construct a parallel to the other, the lines constructed from their points of intersection to the midpoints of the sides divide the quadrilateral into four equal parts. 81. Suppose that CE, DF are perpendiculars to the diameter of a semicir- cle, and two circles are constructed touching CE, DE, and the semicircle, one internally and the other externally. Prove that the rectangle contained by the perpendiculars from their centers on AB is equal to CE.DF. 82. If segments are constructed from any point in the circumference of a circle to the angular points of any inscribed regular polygon of an odd number of sides, the sums of the alternate lines are equal. 83. If at the endpoints of a chord constructed through a given point within a given circle tangents are constructed, the sum of the reciprocals of the per- pendiculars from the point upon the tangents is constant. 84. If a cyclic quadrilateral is such that three of its sides pass through three fixed collinear points, the fourth side passes through a fourth fixed point, collinear with the three given ones. 85. If all the sides of a polygon are parallel to given lines and if the loci of all the angles except for one are segments, the locus of the remaining angle is also a segment. 6.2. PROPOSITIONS FROM BOOK VI 281 86. If the vertical angle and the bisector of the vertical angle is given, the sum of the reciprocals of the containing sides is constant. 87. If P, P’ denote the areas of two regular polygons of any common num- ber of sides inscribed and circumscribed to a circle, and II, IT are the areas of the corresponding polygons of double the number of sides, prove that II is a geometric mean between P and P’ and II 7 a harmonic mean between II and P. 88. The difference of the areas of the triangles formed by joining the cen- ters of the circles constructed about the equilateral triangles constructed out- wards on the sides of any triangle is equal to the area of that triangle. Prove the same if they are constructed inwards. 89. In the same case as #88, the sum of the squares of the sides of the two new triangles is equal to the sum of the squares of the sides of the original triangle. 90. Suppose that R, r denote the radii of the circumscribed and inscribed circles to a regular polygon of any number of sides, R’, r’, corresponding radii to a regular polygon of the same area, and double the number of sides. Prove 91. If the altitude of a triangle is equal to its base, the sum of the distances of the orthocenter from the base and from the midpoint of the base is equal to half the base. 92. In any triangle, when the base and the ratio of the sides are given, the radius of the circumscribed circle is to the radius of the circle which is the locus of the vertex as the difference of the squares of the sides is to four times the area. 93. Given the area of a parallelogram, one of its angles, and the difference between its diagonals, construct the parallelogram. 94. If a variable circle touches two equal circles, one internally and the other externally, and perpendiculars fall from its center on the transverse tan- gents to these circles, the rectangle of the intercepts between the feet of these perpendiculars and the intersection of the tangents is constant. 95. Given the base of a triangle, the vertical angle, and the point in the base whose distance from the vertex is equal half the sum of the sides, con- struct the triangle. 96. If the midpoint of the base BC of an isosceles triangle AABC is the center of a circle touching the equal sides, prove that any variable tangent to the circle will cut the sides in points D, E, such that the rectangle BD.CE is constant. 97. Inscribe in a given circle a trapezium, the sum of whose opposite par- allel sides is given and whose area is given. 6.2. PROPOSITIONS FROM BOOK VI 282 98. Inscribe in a given circle a polygon all of whose sides pass through given points. 99. If two circles oABC, oXYZ are related such that a triangle may be in- scribed in oABC and circumscribed about oXYZ, prove that an infinite num- ber of such triangles can be constructed. 100. In the same case as #99: the circle inscribed in the triangle formed by joining the points of contact on oXYZ touches a given circle. 101. In the same case as #99: the circle constructed about the triangle formed by drawing tangents to oABC at the angular points of the inscribed triangle touches a given circle. 102. Find a point, the sum of whose distances from three given points is a minimum. 103. A line constructed through the intersection of two tangents to a circle is divided harmonically by the circle and the chord of contact. 104. Construct a quadrilateral similar to a given quadrilateral whose four sides pass through four given points. 105. Construct a quadrilateral similar to a given quadrilateral whose four vertices lie on four given lines. 106. Given the base of a triangle, the difference of the base angles, and the rectangle of the sides, construct the triangle. 107. Suppose that [HABCD is a square, the side CD is bisected at E, and the line EF is constructed making the angle ZAEF = ZEAB. Prove that EF divides the side BC in the ratio of 2 : 1. 108. If any chord is constructed through a fixed point on a diameter of a circle, its endpoints are joined to either end of the diameter, and the joining lines cut off on the tangent at the other end, then the portions whose rectangle is constant. 109. If two circles touch and through their point of intersection two secants be constructed at right angles to each other, cutting the circles respectively in the points A, A’ B, B’ then AA’ 2 + BB’ 2 is constant. 110. If two secants stand at right angles to each other which pass through one of the points of intersection of two circles also cut the circles again, and the line through their centers is the two systems of points a, b, c; a’, b’, d respectively, then ab : be :: a’b’ : b’c’ . 111. If a chord of a given circle stands opposite a right angle at a given point, the locus of the intersection of the tangents at its endpoints is a circle. 112. The rectangle contained by the segments of the base of a triangle made by the point of intersection of the inscribed circle is equal to the rectangle 6.2. PROPOSITIONS FROM BOOK VI 283 contained by the perpendiculars from the endpoints of the base on the bisector of the vertical angle. 113. If O is the center of the inscribed circle of the triangle, prove OA 2 OB 2 OC 2 _ be ca ab 114. State and prove the corresponding theorems for the centers of the escribed circles. 115. Suppose that four points A, B, C, D are collinear. Find a point P at which the segments AB, BC, CD stand opposite equal angles. 116. The product of the bisectors of the three angles of a triangle whose sides are a, b, c, is 8a6cs.area (a + b)(b + c)(c + a) 117. In the same case as #116, the product of the alternate segments of the sides made by the bisectors of the angles is a 2 b 2 c 2 (a + b)(b + c)(a + c) 118. If three of the six points in which a circle meets the sides of any trian- gle are such that the lines joining them to the opposite vertices are concurrent, the same property is true of the three remaining points. 119. If a triangle \/A’B’C is inscribed in another AABC, prove AB’.BC’.CA’ + A’B.B’C.C’A is equal to twice the triangle \/A!B’C multiplied by the diameter of the circle oABC. 120. Construct a polygon of an odd number of sides being given that the sides taken in order are divided in given ratios by fixed points. 121. If the external diagonal of a quadrilateral inscribed in a given circle is a chord of another given circle, the locus of its midpoint is a circle. 122. If a chord of one circle is a tangent to another, the line connecting the midpoint of each arc which it cuts off on the first to its point of intersection with the second passes through a given point. 123. From a point P in the plane of a given polygon, suppose that perpen- diculars fall on its sides. If the area of the polygon formed by joining the feet of the perpendiculars is given, the locus of P is a circle. 124. The medians of a triangle divide each other in the ratio of 2 : 1. CHAPTER 7 Infinite Primes This brief chapter highlights Proposition IX. 20, Euclid’s proof of infinitely many prime numbers. 7.1. Definitions 1. The set of all positive integers (1, 2, 3, …) is called the set of natural numbers. 2. A natural number that has only one pair of factors, namely 1 and itself, is a prime number. 3. A natural number that has more than one pair of factors is a composite number. 4. The only natural number that is neither prime nor composite is 1. 7.2. The Proposition This proof is based on Chris K. Caldwell’^] proof which is in turn based on Euclid’s originaQ PROPOSITION 7.1. Any list of prime numbers which is finite is incomplete. PROOF. Suppose we obtain a list of all prime numbers, and the list is finite. Call the primes in our finite list p\u00b1, p 2 , p r – Let P be any common multiple of these primes plus one (for example, P = pi \u2022 p 2 \u2022 … \u2022 p r + 1). Now P is either prime or it is not. If it is prime, then P is a prime that was not in our list. If P is not prime, then it is divisible by some prime: call it p. Notice p cannot be any of pi, p 2 , …,p r , otherwise p would divide 1, a contradiction since the natural numbers do not contain fractions. So this prime p is some prime that was not in our original list. In either case, the original finite list was incomplete, and hence the actual list must be infinite in length. \u25a1 _caldwell@utm.edu ^http : \/\/primes .utm.edu\/notes\/proofs\/infinite\/euclids .html 284 CHAPTER 8 Planes, coplanar lines, and solid angles This chapter includes the first 21 propositions from Euclid’s Book XL 8.1. Definitions 1. When two or more lines are in one plane, they are said to be coplanar. 2. The angle which one plane makes with another is called a dihedral angle. 3. A solid angle is that which is made by more than two plane angles in different planes which meet a point. 4. The point from [Def. 7.3] is called the vertex of the solid angle. 5. If a solid angle is composed of three plane angles, it is called a trihedral angle if of four, a tetrahedral angle; and if of more than four, a polyhedral angle. 6. A line which is perpendicular to a system of concurrent and coplanar lines is said to be perpendicular to the plane of these lines and is also called normal to it. (These lines will sometimes be called “normals” to a given plane. We may also have rays and segments which are normal to a plane.) 7. If from every point in a given line normals are drawn to a given plane, the locus of their feet is called the projection of the given line on the plane. 8. Two planes which meet are perpendicular to each other when the lines constructed perpendicular in one of them to their common section are normals to the other. 9. When two planes which meet are not perpendicular to each other, their inclination is the acute angle contained by two lines drawn from any point of their common section at right angles to it (one in one plane, and one in the other). 10. If at the vertex O of a trihedral angle O \u2014 ABC we construct normals OA, OB, OC to the faces OBC, OCA, OAB, respectively, in such a way that OA is on the same side of the plane OBC as OA, etc., the trihedral angle O \u2014 A’B’C is called the supplementary of the trihedral angle O \u2014 ABC. 285 8.2. PROPOSITIONS FROM BOOK XI: 1-21 286 8.2. Propositions from Book XI: 1-21 PROPOSITION 8.1. If part of a line stands on a plane, then each part of that line must stand on that plane. PROOF. Construct the line AB on the plane X and cut AB at point C. We wish to show that BC is also on plane X. Figure 8.2.1. [7.1] Since AB is on the plane X, it can be extended on X [Postulate 1.2]. Extend it to D. Then, if BC is not on X, let any other plane passing through AD be turned round AD until it passes through the point C. Because the points B, C are in this second plane, the line BC is on it. Therefore, the two lines ABC, ABD lying in one plane have a common segment AB, a contradiction. \u25a1 COROLLARY. 1. [7.1] holds for rays, mutatis mutandis. PROPOSITION 8.2. Two segments which intersect one another at any point are coplanar as are any three segments which form a triangle. PROOF. We claim that two segments (AB, CD) which intersect one another at a point (E) are coplanar as are any three segments (EC, CB, BE) which form a triangle. Figure 8.2.2. [7.2] 8.2. PROPOSITIONS FROM BOOK XI: 1-21 287 Let any plane pass through EB and be turned round it until it passes through C. Then because the points E, C are in this plane, the segment EC is in it [Def. 1.6]. For the same reason, the segment BC is on it. Therefore, EC, CB, BE are coplanar. Since AB and CD are two of these segments, AB and CD are coplanar. \u25a1 PROPOSITION 8.3. If two planes cut one another, their intersection is a line. PROOF. We claim that if two planes (AB, BC) cut one another, their inter- section is a line (BD). 1 K Figure 8.2.3. [7.3] Otherwise, construct in the plane AB the line BED and in the plane BC construct the line BED. Then the lines BED, BED enclose a space, which contradicts [Axiom 1.10]. Therefore, the common section BD of the two planes is a line. \u25a1 PROPOSITION 8.4. If a line is perpendicular to each of two intersecting lines, it will be perpendicular to any line which is both coplanar and concurrent with the intersecting lines. PROOF. If a line (EE) is perpendicular to each of two intersecting lines (AB, CD), it will be perpendicular to any line (GH) which is both coplanar and concurrent with them. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 288 Through any point G in GH construct a line BC intersecting AB, CD which is bisected at G. Join any point F in EF to B, G, C. Then because EF _L EB and EF _L EC, we have that BF 2 = BE 2 +EF 2 and CF 2 = CE 2 + EF 2 BF 2 + CF 2 = BE 2 + CE 2 + 2EF 2 Also, \u00a3F 2 + FF 2 = 2\u00a3G 2 + 2GF 2 [2.10, Ex. 2], and BE 2 + CE 2 = 2BG 2 + 2GE 2 => 2BG 2 + 2GF 2 = 2FG 2 + 2GF 2 + 2FF 2 GF 2 = GF 2 + FF 2 Hence, the angle ZGEF is right, and so FF _L FG. \u25a1 COROLLARY. 1. The normal is the least line that may be constructed from a given point to a given plane; of all others that may be constructed to it, the lines of any system making equal angles with the normal are equal to each other COROLLARY. 2. A perpendicular to each of two intersecting lines is normal to their plane. PROPOSITION 8.5. If three concurrent lines have a common perpendicular, they are coplanar 8.2. PROPOSITIONS FROM BOOK XI: 1-21 289 PROOF. If three concurrent lines (BC, BD, BE) have a common perpen- dicular (AB), they are coplanar. Suppose that BC is not coplanar with BD, BE, and let the plane of AB, BC intersect the plane of BD, BE at the line BF. By [7.3], BE is a segment; and, since it is coplanar with BD, we have that BE is perpendicular to AB (since each are perpendicular to AB, BF) [7.4]. Therefore, the angle ZABF is right. We also have that the angle ZABC is right by hypothesis. Hence, ZABC = ZABF, a contradiction [Axiom 1.9]. Therefore, the lines BC, BD, BE are coplanar. \u25a1 PROPOSITION 8.6. TWO NORMAL LINES. If two lines are normals to the same plane, they are parallel to one another PROOF. If two lines (AB, CD) are normals to the same plane (X), then AB || CD. Figure 8.2.5. [7.5] Figure 8.2.6. [7.6] 8.2. PROPOSITIONS FROM BOOK XI: 1-21 290 Let AB, CD meet the plane X at the points B, D. Join BD, and in the plane X construct DE _L BD. Take any point E in DE. Join BE, AE, AD. Then because AB is normal to X, the angle A ABE is right. Because the angle ZBDE is right, it follows that AE 2 = AB 2 + \u00a3\u00a3 2 = AB 2 + \u00a3L> 2 + L>\u00a3 2 But AB 2 + #L> 2 = AD 2 because the angle ZABD is right. Hence AE 2 = AD 2 + L>\u00a3; 2 . Therefore the angle ZADE is right [1.48]. And since CD is nor- mal to the plane X, DE JL CD. Hence DE is a common perpendicular to the three concurrent lines CD, AD, BD. Therefore these lines are coplanar [7.5]. But AB is coplanar with AD, BD [7.2]. Therefore the lines AD, BD, CD are coplanar; and since the angles ZABD, ZBDC are right, we have that AB || CD [1.28]. \u25a1 Exercises. 1. The projection of any line on a plane is a straight line. 2. The projection on either of two intersecting planes of a normal to the other plane is perpendicular to the line of intersection of the planes. Proposition 8.7. PARALLEL LINES AND THEIR INTERSECTIONS. Two parallel lines and any line intersecting them are coplanar. PROOF. Two parallel lines (AB, CD) and any line (EF) intersecting them are coplanar. If possible, let the intersecting segment lie outside of the plane as segment EGF. In the plane, construct the segment EHF. Then we have two segments EGF, EHF enclosing a space, which contradicts [Axiom 1.10]. Hence, the two parallel straight lines and the transversal are coplanar. \u25a1 Alternatively: 8.2. PROPOSITIONS FROM BOOK XI: 1-21 291 PROOF. Since the points E, F are in the plane of the parallels, the segment joining these points also lies in that plane. \u25a1 PROPOSITION 8.8. NORMAL PARALLEL LINES. If one of two parallel straight lines is normal to a plane, the other line is normal to the same plane. PROOF. If one of two parallel straight lines (AB, CD) is normal to a plane (X), then the other line is normal to the same plane. Figure 8.2.8. [7.8] Let AB, CD meet in the plane X at the points B, D. Join BD. Then the lines AB, BD, CD are coplanar. Wlog, suppose that AB is normal to the X. Construct DE _L BD. Take any point E in DE and join BE, AE, AD. Then because AB is normal to the plane X, it is perpendicular to the line BE in that plane [Def 7.6]. Hence, the angle ZABE is right, and so AE 2 = AB 2 + BE 2 = AB 2 + BD 2 + DE 2 (see the proof of [7.6]). Therefore, the angle ZADE is right. Hence, DE is at right angles both to AD and BD. Therefore DE _L CD [7.4], and DE is coplanar and concurrent with AD and BD. Again, since AB || CD, ZABD + ZBDC is two right angles [1.29]. Since ZABD is right by hypothesis, it follows that ZBDC is also right. Hence CD is perpendicular to the two lines DB, DE, and therefore it is normal to the plane X by [7.4]. \u25a1 PROPOSITION 8.9. TRANSITIVITY OF PARALLEL LINES. Two lines which are each parallel to a third line are also parallel to one another. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 292 PROOF. Two lines (AB, CD) which are each parallel to a third line (EF) are also parallel to one another. If the three lines are coplanar, the proposition is evidently the same as [1.30]. Otherwise, from any point G in EF, construct in the planes of EF, AB and EF, CD (respectively) the lines GH, GK where each is perpendicular to EF [1.11]. Because EF is perpendicular to each of the lines GH, GK, it is normal to their plane [7.4]. And because AB \\ EF by hypothesis and EF is normal to the plane GHK, AB is normal to the plane GHK [7.8]. Similarly, CD is normal to the plane HGK. Hence, since AB and CD are normals to the same plane, they are parallel to one another. \u25a1 PROPOSITION 8.10. ANGLES AND PARALLEL LINES. If two intersecting lines are respectively parallel to two other intersecting straight lines, the angle between the former is equal to the angle between the latter PROOF. If two intersecting lines (AB, BC) are respectively parallel to two other intersecting lines (DE, EF), the angle (ZABC) between the former is equal to the angle (ZDEF) between the latter. Figure 8.2.9. [7.9] i Figure 8.2.10. [7.10] 8.2. PROPOSITIONS FROM BOOK XI: 1-21 293 If both pairs of lines are coplanar, the proposition is the same as [1.29, #2]. Otherwise, take any points A, C in the lines AB, BC and cut off ED = BA, and EF = BC 1.3]. Join AD, BE, CF, AC, DF. Then because AB is equal and parallel to DE, AD is equal and parallel to BE [1.33]. Similarly, CF = BE and CF _L BE. Hence, AD = CF, AD || CF [7.9], and AC = DF [1.33]. Therefore, the triangles AABC, ADEF have the three sides of one respectively equal to the three sides of the other. By [1.8], ZABC = ADEF. \u25a1 Proposition 8.11. CONSTRUCTION OF A NORMAL LINE I. We wish to construct a normal to a given plane from a given point not in the plane. PROOF. We wish to construct a normal to a given plane (BH) from a given point (A) not in the plane. In the given plane BH construct any line BC, and from A construct AD _L BC [1.12]; if AD is perpendicular to the plane, the proof follows. Otherwise, from D construct DE in the plane BH at right angles to BC [1.11] and from A construct AF _L DE [1.12]. We claim that AF is normal to the plane BH. To see this, construct GH \\ BC. Because BC is perpendicular both to ED and DA, it is normal to the plane of ED, DA [11.4]. And since GH \\ BC, it is normal to the same plane [11.8]. Hence AF _L GH [Def. 11.6], and AF _L DE by construction. Therefore, AF is normal to the plane of GH and ED, which is the plane BH. \u25a1 Figure 8.2.11. [7.11] PROPOSITION 8.12. CONSTRUCTION OF A NORMAL LINE II. Construct a normal to a given plane from a given point in the plane. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 294 PROOF. We wish to construct a normal to a given plane from a given point (A) in the plane. Figure 8.2.12. [7.12] From any point B not in the plane construct BC normal to it [11.11]. If this line passes through A, it is the normal required. Otherwise, from A construct AD \\ BC [1.32]. Because AD \\ BC and BC is normal to the plane, AD is also normal to the plane [11.8], and it is drawn from the given point. Hence, the proof. \u25a1 PROPOSITION 8.13. UNIQUENESS OF NORMAL LINES. From a given point, there exists a unique normal to a given plane. PROOF. From a given point (A), we claim that there exists a unique normal to a given plane (X). Figure 8.2.13. [7.13] We shall prove this claim in two parts: 1. Let A is in the given plane and suppose that AB, AC are both normals to it on the same side. Let the plane of BA, AC cut the given plane X at the line DE. Because BA is a normal, the angle ZBAE is right. Similarly, ZCAE is right. Hence ZBAE \u2014 ZCAE, a contradiction. 2. If the point A is above the plane, there can exist only one normal; other- wise, the two would be parallel to one another [11.6], a contradiction. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 295 Hence, the proof. \u25a1 PROPOSITION 8.14. PARALLEL PLANES. Planes which have a common normal are parallel to each other PROOF. We claim that planes (CD, EF) which have a common normal (AB) are parallel to each other. Let GH be the line of intersection between planes CD, EF. Join AK, BK. Because AB is normal to the plane CD, AB JL AK, which it meets in that plane [Def. 11.6], the angle ZBAK is right. Similarly, the angle ZABK is right, and the plane triangle AABK has two right angles, a contradiction. Since the Exercises. 1. The angle between two planes is equal to the angle between two inter- secting normals to these planes. 2. If a line is parallel to each of two planes, the sections which any plane passing through the line makes with the planes are parallel. 3. If a line is parallel to each of two intersecting planes, it is parallel to their intersection. 4. If two lines are parallel, they are parallel to the common section of any two planes passing through them. i Figure 8.2.14. [7.14] planes CD, EF do not intersect, they are parallel. \u25a1 8.2. PROPOSITIONS FROM BOOK XI: 1-21 296 5. If the intersections of several planes are parallel, the normals drawn to them from any point are coplanar. Proposition 8.15. CHARACTERISTIC OF PARALLEL PLANES. Two planes are parallel if two intersecting lines on one plane are respectively parallel to two intersecting lines on the other plane. PROOF. Two planes {AC, DF) are parallel if two intersecting lines (AB, BC) on one plane are respectively parallel to two intersecting lines (DE, EF) on the other plane. From B construct BG perpendicular to the plane DF [11.11] and let it intersect plane DF at point G. Through G construct GH \\ ED and GK \\ EF. Since GH \\ ED by construction and AB \\ ED by hypothesis, AB \\ GH [11.9]. Hence, ZABG + ZBGH equals two right angles [1.29]. Since ZBGH is a right angle by construction, ZABG is also right. Similarly, ZCBG is right. Hence BG is normal to the plane AC [Def. 11.6] as well as normal to DF by construction. Hence the planes AC, DF have a common normal BG; therefore, they are parallel to one another. \u25a1 Proposition 8.16. PARALLEL PLANES AND AN INTERSECTING PLANE. If two parallel planes are cut by a third plane, their common sections with the third plane are parallel. PROOF. If two parallel planes (AB, CD) are cut by a third plane (FG), their common sections with the third plane (EF, GH) are parallel. F Figure 8.2.15. [7.15] 8.2. PROPOSITIONS FROM BOOK XI: 1-21 297 Figure 8.2.16. [7.16] If the lines EF, GH are not parallel, they must meet at some finite dis- tance. Let them meet at K. Since K is a point on the line EF and EF is on the plane AB, K is in the plane AB. Similarly, K is a point on the plane CD. Hence, the planes AB, CD meet at K, a contradiction since they are parallel. Therefore, the lines EF, GH are parallel. \u25a1 Exercises. 1. Parallel planes intercept equal segments on parallel lines. 2. Parallel lines intersecting the same plane make equal angles with that plane. 3. A straight line intersecting parallel planes makes equal angles with he parallel planes. Proposition 8.17. PROPORTIONAL AND PARALLEL LINES. If two parallel lines are cut by three parallel planes in two triads of points, their seg- ments between those points are proportional. PROOF. If two parallel lines (AB, CD) are cut by three parallel planes (GH, KL, MN) in two triads of points (A, E, B and C, F, D where A, E, B and C, F, D are respectively collinear), then the segments between those points are proportional; or, AE : EB :: CF : FD. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 298 Figure 8.2.17. [7.17] Join AC, BD, AD. Let AD meet the plane KL at point X. Join EX, XF. Because the parallel planes KL, MN are cut by the plane ABD at the lines EX, BD, these lines are parallel [11.16]. Hence AE : EB :: AX : XD [6.2]. Similarly, AX : XD :: CF : FD. By [5.11], it follows that AE : EB :: CF : FD. \u25a1 Proposition 8.18. TRANSITIVITY OF PERPENDICULAR PLANES. If a line is normal to a plane, any plane passing through the line is perpendicular to that plane. PROOF. If a line (AB) is normal to a plane (CI), we claim that any plane (DE) passing through the line is perpendicular to CI. Figure 8.2.18. [9.18] Let CE be the common section of the planes DE, CI. From any point F on CE, construct FG in the plane DE such that FG \\ AB [1.31]. Now AB \\ FG and AB is normal to the plane CI; hence, FG is normal to CI [11.8]. Since FG || AB, we have that ZABF + ZBFG are equal to two right angles [1.29]. Since ZABF is right by hypothesis, ZBFG is right and therefore FG _L CE. Hence every line in the plane DE drawn perpendicular to the common section 8.2. PROPOSITIONS FROM BOOK XI: 1-21 299 of the planes DE, CI is normal to the plane CI. Therefore by [Def. 8.11], the planes DE, CI are perpendicular to each other. \u25a1 PROPOSITION 8.19. INTERSECTING PLANES. If two intersecting planes are each perpendicular to a third plane, their common section is normal to that plane. PROOF. If two intersecting planes (AB, BC) are each perpendicular to a third plane {ADC), their common section (BD) is normal to ADC. Figure 8.2.19. [7.19] Otherwise, construct the line DE from D in the plane AB such that DE _L AD where AD is the common section of the planes AB, ADC. In the plane BC, construct BF perpendicular to the common section DC of the planes BC, ADC. Because the plane AB is perpendicular to ADC, the line DE in AB is normal to the plane ADC [Def. 7.8]. Similarly, DF is normal to it. Therefore from the point D there are two distinct normals to the plane ADC, a contradiction [7.13]. Hence, BD is normal to the plane ADC. \u25a1 Exercises. 1. If three planes have a common line of intersection, the normals drawn to these planes from any point of that line are coplanar. 2. If two intersecting planes are respectively perpendicular to two inter- secting lines, the line of intersection of the former is normal to the plane of the latter. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 300 3. In the last case, show that the dihedral angle between the planes is equal to the rectilinear angle between the normals. PROPOSITION 8.20. TRIHEDRAL ANGLES. The sum of any two plane an- gles of a trihedral angle is greater than the third. PROOF. The sum of any two plane angles (ZBAD, ZD AC) of a trihedral angle (at point A) is greater than the third (ZBAC). Figure 8.2.20. [7.20] If the third angle ZBAC is less than or equal to either of the other angles, the proposition is evident. Otherwise, suppose it greater: take any point D in AD and at the point A in the plane BAC make ZBAE = ZBAD [1.23]. Cut off AE = AD. Through E construct BC, cutting AB, AC at the points B, C. Join DB, DC. Then the triangles ABAD, ABAE have the two sides BA, AD in one equal respectively to the two sides BA, AE in the other and ZBAD = ZBAE. Therefore the third side BD = BE. However, the sum of the sides BD, DC is greater than BC; hence DC is greater than EC. Again, because the triangles AD AC, AEAC have the sides DA, AC respectively equal to the sides EA, AC in the other where the base DC greater than EC. By [1.25], the angle ZD AC is greater than ZEAC, but ZDAB = ZBAE by construction. Hence, ZBAD + ZD AC > ZBAC. \u25a1 PROPOSITION 8.21. SUM OF PLANE ANGLES. The sum of all the plane angles forming any solid angle is less than four right angles. PROOF. The sum of all the plane angles (ZBAC, ZCAD, etc.) forming any solid angle (at A) is less than four right angles. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 301 Figure 8.2.21. [7.21] Suppose for the sake of simplicity that the solid angle at A is contained by five plane angles ZBAC, ZCAD, ZDAE, ZEAF, ZFAB. Let the planes of these angles be cut by another plane at the lines BC, CD, DE, EF, FB. By [11.20], we have ZABC + ZABF greater in measure than ZFBC, ZACB + ZACD greater in measure than ZBCD, etc. Adding these, we obtain the sum of the base angles of the five triangles ABAC, ACAD, etc., greater than the sum of the interior angles of the penta- gon BCDEF; that is, greater than six right angles. But the sum of the base angles of the same triangles, together with the sum of the plane angles ZBAC, ZCAD, etc., forming the solid angle at A is equal to twice as many right angles as there are triangles ABAC, ACAD, etc.; that is, equal to ten right angles. Hence, the sum of the angles forming the solid angle is less than four right angles. \u25a1 Observation: this proposition may not hold if the polygonal base BCDEF contains re-entrant angles. Chapter 7 exercises. 1. Any face angle of a trihedral angle is less than the sum (but greater than the difference) of the supplements of the other two face angles. 2. A solid angle cannot be formed of equal plane angles which are equal to the angles of a regular polygon of n sides except when n = 3, 4, 5. 3. Through one of two non-coplanar lines, construct a plane parallel to the other. 4. Construct a common perpendicular to two non-coplanar lines and show that it is the shortest distance between them. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 302 5. If two of the plane angles of a tetrahedral angle are equal, the planes of these angles are equally inclined to the plane of the third angle, and conversely. If two of the planes of a trihedral angle are equally inclined to the third plane, the angles contained in those planes are equal. 6. Prove that the three lines of intersection of three planes are either par- allel or concurrent. 7. If a trihedral angle O is formed by three right angles and A, B, C are points along the edges, the orthocenter of the triangle AABC is the foot of the normal from O on the plane ABC. 8. If through the vertex O of a trihedral angle O \u2014 ABC any line OD is drawn interior to the angle, the sum of the rectilinear angles ADO A, ZDOB, ZDOC is less than the sum but greater than half the sum of the face angles of the trihedral. 9. If on the edges of a trihedral angle O \u2014 ABC three equal segments OA, OB, OC are taken, each of these is greater than the radius of the circle de- scribed about the triangle AABC. 10. Given the three angles of a trihedral angle, find by a plane construction the angles between the containing planes. 11. If any plane P cuts the four sides of a Gauche quadrilateral ABCD (a quadrilateral whose angular points are not coplanar) at four points, a, b, c, d, then given the four ratios Aa Bb Cc Dd aB’ bC’ cD 9 dA we have that Aa Bb Cc Dd aB ‘ bC ‘ cD ‘ ~dA ~ Conversely, if Aa Bb Cc Dd _ , aB ‘ bC ‘ cD ‘ ~dA ~ then the points a, b, c, d are coplanar. 12. If in #11 the intersecting plane is parallel to any two sides of the quadrilateral, it cuts the two remaining sides proportionally. 13. IfO\u2014A’B’C is the supplementary of O \u2014 ABC, prove that O \u2014 ABC is the supplementary angle of O \u2014 A’B’C 14. If two trihedral angles are supplementary, each dihedral angle of one is the supplement of the corresponding face angle of the other. 15. Through a given point, construct a line which will meet two non- coplanar lines. 16. Construct a line parallel to a given line which will meet two non- coplanar lines. 8.2. PROPOSITIONS FROM BOOK XI: 1-21 303 17. Given an angle ZAOB, prove that the locus of all the points P of space where the sum of the projections of the line OP on OA and OB are constant is a plane. Part 3 Student Answer Key CHAPTER 1 Solutions: Angles, Parallel Lines, Parallelograms [1.1] Exercises The following two exercises use Fig. 1.5.1 (above) and are to be solved when the student has completed Chapter 1. 1. If the segments AF, BF are joined, prove that the figure [HACBF is a lozenge. PROOF. Suppose that AF, BF are joined. By an argument similar to the proof of [1.1], we have that AB = AF = BF. By [1.1], we have that AC = AB = BC. Hence, it follows that AC = BC = BF = AF and so by [Def 1.29], BACBF is a lozenge. \u25a1 2. If AB is extended to the circumferences of the circles (at points D and E), prove that the triangles ACDF and ACEF are equilateral. PROOF. Construct segments CD, DF, FE, CE, and CF as per the hypoth- esis. Also extend AB to a line. We wish to show that ACDF and ACEF are equilateral. 305 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 306 Figure 1.0.1. [1.1, #2] By [1.1], ZACB = ZBAC = ZABC. [1.32, Cor. 6] states that each angle of an equilateral triangle equals two- thirds of a right angle; since a right angle equals \u00a7 radians, ZACB = ZBAC = ZABC = f . Since [HACBF is a lozenge by [1.1, #1], CF is its Axis of Symmetry. Hence, ZACL = f . Consider AACL. It follows that ZALC = f . Similarly, we can show that in ABCL ZBLC = f , ZBCL = f , and ZLBC = \u00a7 . Since AACL and ABCL share side CL, by [1.26] we have that AACL ^ ABCL. Consider AD AC. Notice that ZD AC = ?f since ZD AC and ZLAC are supplementary. Since AC and AD are both radii of oCDF, we have that ZADC = ZACD = | [1.5]. Hence in ADCL, ZLDC = \u00a7 and ZDCL = f . Similarly, we can show that in AECL that ZLEC = f and ZLCE = f . Since ADCL and A\u00a3CL share side CL, by [1.26] we have that ADCL ^ AECL. It follows that DC = CE Similarly we can show that ADFL ^ AEFL, and so DF = FE. This will also show that ZFDL = ZCDL, ZFLD = ZCLD. Since ACDL and AFDL share side DL, by [1.26], ACDL ^ AFDL, so DC = DF. Hence, DC = CE = DF = FE Finally, we have that ZFDC = ZFEC, ZDCF = ZECF, and ADCF and AECF share side CF. By [1.26], ACDF ^ ACEF. Since each angle of each triangle equals | radians, the triangles are each equilateral. \u25a1 Corollary, bcefd is a lozenge. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 307 [1.2] Exercises 1. Prove [1.2] when A is a point on BC. PROOF. Let BC be an arbitrary segment such that A is a point on BC. If A = BorA = C, the proof follows trivially. Suppose that A is not an endpoint of BC. Construct the equilateral trian- gle AABD. Also construct the circle oCEF with center A and radius equal in length to AC. Extend side DA to the point E on the circumference of oCEF. Figure 1.0.2. [1.2, #1] Since AABD is equilateral, AB = AD. Since AC and AE are radii of oCEF, AC = AE. Hence, DE = AD \u00a9 AE = BC By using [Axiom 1.8] (specifically the principle of superposition), we may move DE so that the point D coincides with point A. The proof follows. \u25a1 [1.4] Exercises Prove the following: 1. The line that bisects the vertical angle of an isosceles triangle bisects the base perpendicularly. PROOF. Suppose that AABC is an isosceles triangle (where AB = AC). Further suppose that the ray AD bisects the angle ABAC. (A line, ray, or a 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 308 segment of appropriate size may be used in this proof, mutatis mutandis.) We wish to show that ZADB = ZADC and that BD = CD. Figure 1.0.3. [1.4, #1] Since AB = AC and ZDAB = ZD AC by hypothesis, and since AABD and AACD share the side AD, by [1.4] we have that AABD ^ AACD Hence, BD = CD. Also, ZADB = ZADC. Since the ZADB and ZADC are supplements, they are right angles by [Def. 1.14]; hence, AD is a perpendicular bisector of the base (AC) of A ABC. \u25a1 2. If two adjacent sides of a quadrilateral are equal and the diagonal bi- sects the angle between them, then their remaining sides are equal. PROOF. Suppose that ABCD is a quadrilateral where AB = AC and where the diagonal AD bisects ZBAC. We wish to show that BD = CD. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 309 Figure 1.0.4. [1.4, #2] Since AC = AB and ZCAD = ZDAB by hypothesis, and since AACD and AABD share the side AD, by [1.4] we have that AACD ^ AABD Hence, BD = CD. \u25a1 3. If two segments stand perpendicularly to each other and if each bisects the other, then any point in either segment is equally distant from the end- points of the other segment. PROOF. Suppose that segments AB and CD stand perpendicularly to each other and bisect each other at point E. Let F be a point on AB. We claim that F is equally distant from C and D. Figure 1.0.5. [1.4, #3] 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 310 Construct ACEF and ADEF. Since AB perpendicularly bisects CD, ZCEF ZDEF and CE = DE. Since ACEF and ADEF share side FE, by [1.4] we have that ACEF ^ ADEF Hence, CF = DE. The proof for a point on CD is similar to the above, mutatis mutandis. \u25a1 [1.5] Exercises 2. Prove that the line joining the point A to the intersection of the segments CF and BG is an Axis of Symmetry of AABC. PROOF. Construct the line AH on the figure from [1.5] where H is in the intersection of the segments CF and BG. We wish to show that AH is the Axis of Symmetry of AABC. Figure 1.0.6. By [1.5], we have that AFBC ^ AGCB. Subtracting AHBC from each, we have that AFBH ^ AGCH. Hence, HB = HC. Since F was chosen arbitrarily, the position of H is arbitrary in the proof of [1.5], and so we have that the distance from B to any point on the line AH is equal to the distance from C to that point. Let \/ be the point on AH which intersects the base of AABC. By the above, BI = CI. By [Def. 1.35], AH is the Axis of Symmetry of AABC. \u25a1 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 311 6. If three points are taken on the sides of an equilateral triangle (one on each side and at equal distances from the angles), then the segments joining them form a new equilateral triangle. PROOF. Suppose that AABC is equilateral. Construct points \/, J, and K on sides BC, AB, and AC, respectively. Since each point on each side is at an equal distance from the endpoints of the relevant side, each point is the midpoint of the side on which it stands. Construct We wish to show that AUK is equilateral. Figure 1.0.7. Since I is the midpoint of side BC, CI = BI. Since J is the midpoint of side AB, J A = JB; since AABC is equilateral, IB = JB. Continuing in this manner, we can show that IB = JB = JA = AK = KC = IC And since AABC is equilateral, by [1.5, Cor. 1] we also have that ZABC = ZACB = ZBAC Hence by [1.4], AJBI = AKCI = AJAK It follows that IJ = JK = KI, and so AUK is equilateral. \u25a1 [1.9] Exercises 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 312 A 2. Prove that AF _L DE. (Hint: the proof follows almost immediately from [1.5, #2].) PROOF. Extend AF to a line. By [1.5, #2], AF is the Axis of Symmetry of AADE. Hence, if H is the intersection of DE and AF, then DH = HE. Consider AADH and AAEH. Since each triangle shares side AH and DA = EA by construction, by [1.8] we have that AADH 9* AADE. Hence, ZL>iF4 = ZEHA. By [Def. 1.14], ZD HA and Z\u00a3iF4 are right angles. Hence, AF _L L>\u00a3. \u25a1 3. Prove that any point on AF is equally distant from the points D and E. PROOF. By [1.5, #2], is an Axis of Symmetry of AABC. The proof of this problem follows immediately by the proof of [1.5, #2]. \u25a1 [1.10] Exercises 1. Bisect a segment by constructing two circles. PROOF. Construct the figure from [1.1]. We wish to bisect segment AB. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 313 Figure 1.0.8. [1.10, #1] If we construct a segment from C which intersects AB and bisects ZACB, the proof follows immediately from the proof of [1.10]. \u25a1 2. Extend CD to a line. Prove that every point equally distant from the points A, B are points in the line CD. PROOF. Construct the figure from [1.10] and extend CD to a line. Figure 1.0.9. [1.10, #2] By [1.9, Cor. 1], the line CD is an Axis of Symmetry to AB. If point E is an equal distance from both points A and B, then E must lie on the Axis of Symmetry of AB, which is CD \u25a1 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 314 [1.11] Exercises 1. Prove that the diagonals of a lozenge bisect each other perpendicularly. PROOF. Construct the figure from [1.11], and also construct the equilateral triangle ADEG where G lies on the opposite side of AB from the point F. Construct the segment GC. By a similar argument to the proof of [1.11], GC _L AB. Consider the triangles ADCF and ADCG. Since DF = DG by construc- tion, the triangles share side DC, and FC = GC by [1.47], we have by [1.8] that ADCF ^ ADCG. Similarly, it follows that ADCF = ADCG = AECF = AECG Hence, FD = FE = GD = GE and so [HFEGD is a lozenge. Clearly, GC 0 CF = GF where GF is a diagonal of BDFEG; similarly, DE is the other diagonal of BDFEG. By the above, GF _L DE. Hence, the proof. \u25a1 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 315 3. Find a point on a given line that is equally distant from two given points. PROOF. Suppose that AB is our given line and that C and D are our given points. We wish to find a point F on AB which is equally distant from C and D. Figure 1.0.11. [1.11, #3] Construct the segment CD and by [1.10] locate its midpoint, E. Construct FE where CD _L FE and F is a point on AB. We claim that F is the required point. Consider ACEF and ADEF. CE = DE by construction, ZCEF = ZDEF by construction, and the triangles share side EF. By [1.4], ACEF 9* ADEF. Hence, CF = DF. \u25a1 5. Find a point that is equidistant from three given points. (Hint: you are looking for the circumcenter of the triangle formed by the points.) PROOF. Construct three arbitrary points A, B, and C. Connect the three points, constructing A ABC. By [1.10], we may locate the midpoints of each side of the triangle where the midpoint of AC is F, the midpoint of BC is E, and the midpoint of AB is D. Construct the line FG such that FG _L AC. Similarly, construct the line EG such that EG _L BD and line DG such that DG JL AB. Since none of the pairs of sides of A ABC are parallel, the lines FG, DG, and EG have a common point of intersection: G. We claim that G is equidistant from A, B, and C. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 316 Construct the segments AG, BG, and CG. Consider AADG and ABDG. Since AD = BD, AB _L DG, and the triangles share side DG, by [1.4], AADG ^ ABDG. Hence, AG = BG. Similarly, consider AAFG and ACFG. By an argument similar to the above, we have that AAFG ^ ACFG, and so AG = CG. It follows that AG = BG = CG and hence G is equidistant from points A, B, C. \u25a1 [1.12] Exercises 1. Prove that circle oFDG cannot meet AB at more than two points. PROOF. Suppose that oFDG intersects AB at more than two points. If the third point lies between points F and G, then the radius of oFDG must decrease in length, a contradiction, since a circle’s radius is a fixed length. Similarly, if the third point lies to the left of F or to the right of G, the radius of oFDG must increase in length, also a contradiction. Hence, oFDG cannot meet AB at more than two points. \u25a1 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 317 [1.19] Exercises 3. Prove that three equal segments cannot be constructed from the same point to the same line. PROOF. Construct the line AB and point C. Construct the segments CA, CB, and CD such that CA = CB. We claim that CD cannot be constructed such that CA = CB = CD. Suppose that CA = CB = CD. By [1.19, Cor. 1], ZCDA = ZCAD and ZCAB = ZCBA. Hence, ZCDA = ZCBD; but by [1.16], we have that ZCDA > ZCBD, a contradiction. A similar result occurs if point D does not fall between points A and B, mutatis mutandis. Hence, the proof. \u25a1 [1.20] Exercises 5. The perimeter of a quadrilateral is greater than the sum of its diagonals. PROOF. Suppose that ABCD is a quadrilateral with diagonals AC and BD. We wish to show that AB + BC + CD + DA > AC + BD 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 318 Figure 1.0.13. [1.20, #5] By [1.20], we have that AD + DC > AC AB + BC > AC AD + AB > BD BC + CD > BD Or, AB + BC + CD + DA > 2 \u2022 AC AB + DC + CD + DA > 2 \u2022 BD 2(AB + BC + CD + DA) > 2(AC + BD) AB + BC + CD + DA > AC + DD \u25a1 6. The sum of the lengths of the three medians of a triangle is less than 3\/2 times its perimeter. PROOF. Construct AABC with medians AF, BE, and CD. We wish to show that AF + BE + CD BE Similarly, in ADBC, we have BD + BC> CD and in AACF, we have AC + CF > AF Adding each inequality, we have that AB + BC + AC + + + CF > AF + BE + CD AB + BC + AC+ AC + ^ AB + ^FC > AF + BE + CD (AB + BC + AC) > AF + BE + CD since AF = \u00b1AC, F\u00a3> = \u00b1 AB, and CF = BC. \u25a1 [1.22] Exercises 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 320 1. Prove that when the above condition is fulfilled (that the sum of every two pairs of segments is greater than the length of the remaining segment) then the two circles must intersect. PROOF. Construct the figure from [1.22] such that the two circles do not intersect where AR = FD = FK, FG = BS, and GH = CT. Figure 1.0.15. [1.22, #1] Consider AFGK. By [1.20], we have that FG + GK > FK FG + GH + HK > FK FG + GH > FK- HK FG + GH < FK BS + CT ZEFD. PROOF. By [1.16], ZBCA > ZBHC, and ZBHC = ZBHA+ZAHC. There- fore, ABC A > ZBHA. By the proof of [1.24], ZBHA = ZEFD, and the proof follows. \u25a1 [1.26] Exercises 1. The endpoints of the base of an isosceles triangle are equally distant from any point on the perpendicular segment from the vertical angle on the base. PROOF. By [1.9, #2] and [1.9, #3], any point on the the perpendicular seg- ment from the vertical angle on the base is equally distant from the endpoints of the base. \u25a1 2. If the line which bisects the vertical angle of a triangle also bisects the base, the triangle is isosceles. PROOF. Construct AABC with segment AD such that ZBAC = ZBAD + ZCAD where ZBAD = ZCAD and where BD = DC. We wish to prove that ZCBA = ZBCA. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 323 Figure 1.0.17. [1.26, #2] By [1.9, #2], AD _L BC. Consider AABD and AACD. Notice that BD = CD, \/BDA = \/LCD A (since both are right angles), and \/BAD = ACAD. By [1.26], AABD ^ AACD. Hence, \/DBA = \/DC A. It follows that \/CBA = \/DBA = \/DC A = \/BCA \u25a1 6. Prove that if two right triangles have equal hypotenuses and that if a side of one is equal in length to a side of the other, then the triangles they are congruent. (Note: this proves the special case of Side-Side-Angle congruency for right triangles.) PROOF. Suppose that AABC and ADEF are right triangles where \/ACB and \/DFE are right angles, the hypotenuses AB and DE are equal in length, and AC = DF. (The proof follows mutatis mutandis if BC = EF.) We wish to show that AABC ^ ADEF. By superposition, we may “move” ADEF so that side AC lies on side DF such that A = D and C = F. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 324 Figure 1.0.18. [1.26, #6] Consider AAEB. Since AE = AB, ZAEB = ZABE by [1.5]. Now consider AAEC and AACB. Since ZACE = ZACB, ZAEC = ZABC, [1.29] Exercises 2. If ZACD, ZBCD are adjacent angles, any parallel to AB will meet the bisectors of these angles at points equally distant from where it meets CD. PROOF. Construct the straight line AB which contains the point C. Con- struct a straight line EE such that AB || EE where D is a point on EE. Construct segments CH and C J such that C J bisects ZACD and CH bisects ZBCD. We wish to show that DH = DJ. Construct segments JK and HL such that JK \\ CD and CD \\ HL. Since CH is a bisector of ZBCD, ZBCH = ZHCD. Since CD \\ HL, ZHCD = ZCHL. Similarly, ZBCH = ZDHC. Hence, and AE = AB, by [1.26], AAEC ^ A ABC. Thus ADEF = AAEC. \u25a1 Figure 1.0.19. [1.29, #2] ZBCH = ZHCD = ZDHC = ZCHL 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 325 z Figure 1.0.20. [1.29, #2] This means that ACDH and ACLH are isosceles triangles where the base angles of each triangle are equal, and so By similar reasoning, ADC J and AKCJ are isosceles triangles where the base angles are equal, and so 5. Two straight lines passing through a point equidistant from two paral- lels intercept equal segments on the parallels. PROOF. Construct straight lines AB and CD such that AB \\ CD. Con- struct straight line LM such that AB JL LM and choose the point G on LM such that GL = GM [1.10]. Construct arbitrary straight lines H J and IK such that each passes through . We claim that HI = JK. CL = HL = CD = DH KJ = CK = CD = DJ Thus, DJ = DH. \u25a1 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 326 Consider AGLI and AGMK: ZLGI = ZMGK by [1.15]; GL = GM by con- struction; ZGMK = ZGLI by construction. Hence by [1.26], AGLI ^ AGMK. It follows that GI = GK. Now consider AGHI and . By [1.15], ZHGI = ZJGK. Since AB \\ CD, by [1.29, Cor. 1], ZGIH = ZGKJ. By the above, GI = GK. Again by [1.26], we have that AGHI ^ AGJK. Thus, HI = JK. \u25a1 [1.31] Exercises 1. Given the altitude of a triangle and the base angles, construct the trian- gle. PROOF. Suppose we are given altitude h (a segment) and base angles a and \/3. Extend h to the line AB where A and B are the endpoints of h. Construct the line BC such that AB _L BC [1.11, Cor. 1]. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 327 Figure 1.0.22. [1.31, #1] If a is the “left angle” of the triangle to be constructed, then \/3 is the “right angle” of the triangle. (Construction is trivial if the left\/right assignment is reversed.) Construct a on the left side of AB as ZBDA such that the ray at point D intersects point A. Similarly, construct f) on the right side of AB as ABC A such that the ray at point C intersects point A. Clearly, figure ACD is a three-sided polygon containing angles a and \/3 whose altitude is h. Hence, AACD is the required triangle. \u25a1 5. Through two given points on two parallel lines, construct two segments forming a lozenge with given parallels. PROOF. Construct parallel lines AB and CD where B and C are our given points. We wish to construct the lozenge FBGC by constructing the segments FC and BG. Figure 1.0.23. [1.31, #5] 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 328 Locate E, the midpoint of BC [1.10]. Construct the line FG such that FG _L BC and where F is a point on AB and G is a point on CD. Construct segments FC and BG. We claim that [HFBGC is the required lozenge. By construction, [HFBGC is a parallelogram. Consider AC\u00a3G and ABFE. By construction, = CE and ZC\u00a3G = Z\u00a3\u00a3F. By [1.29, Cor. 1], ABFE = ZCGE. Then by [1.26], ACEG ^ ABFE, and so BF = CG and \u00a3F = EG. Now consider AF\u00a3C and ABEG. We have that BE = CE, EF = EG, and ZFEC = ZBEG; by [1.4], AFEC ^ A^G. Hence, FC = By [1.32, #6], ZF\u00a3\u00a3 = ZC\u00a3G = ZFEC = ZGEB = right angle By [1.47], BF = BG, and so [1.32] Exercises. 3. If the line which bisects the external vertical angle is parallel to the base, then the triangle is isosceles. PROOF. Construct AABC with external vertical angle ZACE where line CD bisects ZACE and CD \\ AB. We claim that AABC is isosceles. BF = BG = FC = CG Thus by [Def 1.29], FBGC is a lozenge. \u25a1 Figure 1.0.24. [1.32, #3] 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 329 By [1.32], ZACE = ZCAB + ZCBA. By [1.29, Cor. 1], ZDCE = ZCBA; that the same corollary, ZACD = ZCAB. Since ZACD = ZDCE by hypothe- sis, we have that ZACE = 2 \u2022 ZDCE = 2 \u2022 ZACD, from which it follows that ZCAB = ZCBA. By [1.6], AABC is isosceles. \u25a1 An alternate proof that does not use [1.32]: PROOF. Suppose the above. By [1.29, Cor. 1], ZDCE = ZABC. By hypothesis, ZDCE = ZACD, and so ZACD = ZABC. Again by [1.29, Cor. 1], ZACD = ZCAB, and so ZCAB = ZABC. By [1.6], AABC is isosceles. \u25a1 5. The three perpendicular bisectors of a triangle are concurrent. PROOF. Construct AABC, altitudes AG and CF, and the segment BE where E is a point on AC such that AG, BE, and CF intersect at point D. We wish to show that ZAEB is a right angle. Figure 1.0.25. [1.32, #5] Construct segments HD, HI, and AI such that AD = HD = HI = AI By [Def. 1.29], BADHI is a lozenge. By [1.11, #1], the diagonals of a lozenge bisect each other perpendicularly. Since AH is a diagonal of [HADHI, AH is bisected perpendicularly by DI at point E, and so ZAED equals one right angle. If ZEDB equals two right angles, then the proof follows. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 330 By the construction of ZBDA, we have that ZEDB = ZBDA + ZED A and that ZBDA = ZDEA + ZEAD by [1.32]. Hence ZEDB = ZBDA + ZED A = ZDEA + Z\u00a3AD + ZED A = two right angles since ZDEA, ZEAD, and Z\u00a3L>A are the interior angles of AEDA. Thus, ZAED = ZAEB = right angle \u25a1 6. The bisectors of two adjacent angles of a parallelogram are at right angles. PROOF. Construct a parallelogram. If the parallelogram is a lozenge, the result follows from [1.34]. Otherwise, construct parallelogram BABCD which is not a lozenge. Con- struct ZDAF such that ZDAF = ZDAB and ZABE such that ZABE = ZABC. Extend AD to meet BE at G and extend BC to meet AF at H. Con- nect G and H. Since \u00a3C || AD, BR \\ AG; similarly, BA \\ HG. Hence, BABHG is a parallelogram. We claim that EABHG is also a lozenge. x n Figure 1.0.26. [1.32, #6] By hypothesis, ZGAH = ZBAH. By [1.29, Cor. 1], ZGAH = ZAHB and ZBAH = ZAHG. Hence ZGAH = ZBAH = ZAHB = ZAHG 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 331 Consider AABH and AAGH. Since \/LB AH = ZAHB, AABH is isosce- les. A similar argument holds for AAGH. Given the equality of the four base angles above and given than the two triangles share base AH, by [1.26] AABH ^ AAGH. Hence, we have that AB = BH = HG = GA By [Def 1.29], BABHG is a lozenge. \u25a1 [1.33] Exercises 1. Prove that if two segments AB, BC are respectively equal and parallel to two other segments DE, EF, then the segment AC joining the endpoints of the former pair is equal in length to the segment DF joining the endpoints of the latter pair. PROOF. Construct segments AB, BC, DE, EF such that AB = DE, BC = EF, AB || DE, and BC || EF. Construct segments AC and DF. We wish to show AC = DF. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 332 Figure 1.0.27. [1.33, #1] Consider AABC and ADEF. Suppose that ZABC + ZDEF. Move AABC such that C = E. Construct AD such that AD \\ BF. Since AB \\ CD and AD || BF, we have that ZABC = ZDCF. But ZDCF = ZDEF \u00b1 ZABC, a contradiction. Hence, ZABC = ZDEF. By [1.4], AABC ^ ADEF, and so AC = DF. \u25a1 [1.34] Exercises 1. Show that the diagonals of a parallelogram bisect each other. PROOF. Consider BABCD and diagonals AD, BC. Let point E be the intersection of AD, BC. We wish to prove that CE = EB and AE = ED. Since AB \\ CD, ZBCD = ZCBA. Similarly, we have that AC DA = ZDAB. Consider AECD and AAEB. Since ZECD = ZEBA, ZEDC = ZEAB, and CD = AB, by [1.26] we have that AECD 9* AAEB. Hence, AE = ED. A similar argument shows that CE = EB, mutatis mutandis. Since both diagonals are bisected at their point of intersection, the proof follows. \u25a1 2. If the diagonals of a parallelogram are equal, each of its angles are right angles. PROOF. Construct BABCD as in [1.34, #1] and suppose that AD = BC. We wish to show that ZCAB = ZABD = ZBDC = ZACD = 1 right angle By [1.34, #1], we have that EA = ED = EC = EB. Consider AECD, AEDB, AEBA, and AEAC; by the above, they are isosceles triangles. By [1.6, Cor. 1], ZEAC = ZECA = ZECD = ZEDC = ZEDB = ZEBD = ZEBA = ZEAB Consider AABC. Notice that ZECA + ZEAC + ZEAB + ZEBA = 2 right angles 4 \u2022 ZECA = 2 right angles ZECA = right angles By the above equality, it follows that ZCAB = ZABD = ZBDC = ZACD = 1 right angle \u25a1 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 334 6. Construct a triangle being given the midpoints of its three sides. PROOF. Suppose we are given midpoints A, B, and C of a triangle. We wish to construct ADEF such that DA = AE, DC = CF, and EB = BF. Figure 1.0.29. [1.34, #6] Construct A ABC. Also construct line AD such that AD \\ BC, construct line BE such that BE || AC, extend BC to a line, extend AB to a line, construct line CD such that CD || AB, and construct line BD. We claim that ADEF is the required triangle. Notice that if we construct segments EG and FG such that BEGFD is a parallelogram, the point B intersects the two diagonals of ElEGFD. By [1.34, #1], the diagonals are bisected. Hence, EB = BF. By similar constructions, we may show that DA = AE and DC = CF, mutatis mutandis. (The details are left as an exercise to the student.) \u25a1 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 335 [1.37] Exercises 1. If two triangles of equal area stand on the same base but on opposite sides, the segment joining their vertices is bisected by the base. PROOF. Suppose we have AABG and AABI such that both triangles share base AB and point G stands on the opposite side of AB than point \/. We claim that the segment GI is bisected by either AB or by the extension of AB to a line. Figure 1.0.30. [1.37, #1] Extend AB to a line if necessary. Let J be the point where GI intersects AB or its extension. Construct lines GK and LI such that GK \\ AB and LI || AB. By [1.30], GK \\ LI. Also construct lines GB and KI such that GB || KI. Hence, UGKIL is a parallelogram. By [1.34, #1], GJ = JI. Hence, the proof. \u25a1 [1.38] Exercises 1. Every median of a triangle bisects the triangle. PROOF. Construct AABC where AF is the median of side BC. We wish to show that AABF = AACF. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 336 Construct line AE such that AE \\ BC. Clearly, AABF and AACF stand between the same parallels (AE and BC). Since BE = EC by hypothesis, AABF = AACF by [1.38]. \u25a1 Note: we do not claim that the triangles are congruent, merely equal in area. 5. One diagonal of a quadrilateral bisects the other if and only if it also bisects the quadrilateral. PROOF. Suppose we have quadrilateral ABCD with diagonal BD such that BD intersects AC at E such that AE = EC. We claim that AABD = ACBD. Figure 1.0.32. [1.38, #5] Consider AABE and ACBE. By [1.38], they are equal in area; a similar result holds for AAED and AC ED. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 337 Now consider AADB and ACDB. Notice that AADB = AAED \u00a9 AABE = AC ED \u00a9 AC BE = ACDB Now suppose that AABD = ACBD. By [1.37, #1], we have that AE = EC. \u25a1 [1.40] Exercises 1. Triangles with equal bases and altitudes are equal in area. PROOF. Suppose we have two triangles with equal bases and with equal altitudes. Since the altitude of a triangle is the distance between the parallels which contain it, equal altitudes imply that the triangles stand between the same parallels. By [1.38], the triangles are equal in area. \u25a1 2. The segment joining the midpoints of two sides of a triangle is parallel to the third because the medians from the endpoints of the base to these points will each bisect the original triangle. Hence, the two triangles whose base is the third side and whose vertices are the points of bisection are equal in area. PROOF. Construct AABC with midpoint D on side AB and midpoint E on side AC. Connect D and E. We wish to show that DE \\ BC. Figure 1.0.33. [1.40, #2] 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 338 Construct segment DF such that DE = DF. Construct line BF and con- sider AADE and ABDF. Notice that ZADE = ZBDE by [1.15]. Also, since D bisects AB, AD = BD. Finally, DE = DF by construction. Hence by [1.4], AADE ^ ABDF, and so ZFBD = ZDAE. By [1.29, Cor. 1], FB \\ AC. Also, since FB = AE and AE = EC, FB = EC. Since FB and EC are equal in length and parallel, by [1.33] EF and BC are opposite and parallel; hence, BFECB is a parallelogram. Thus, DE \\ BC. \u25a1 COROLLARY. BC = 2 \u2022 DE, or the median segment as constructed is half the length of its opposite and parallel side. 4. The segments which connect the midpoints of the sides of a triangle divide it into four congruent triangles. PROOF. Suppose we have AABC with midpoints D on side BC, E on side AC, and F on side AB. Construct segments DE, EF, and DF. We wish to show that AAEF = AECD = AFDB = ADFE Figure 1.0.34. [1.40, #4] By [1.40, #2], we have that DE || AB, DF || AC, and EF || BC. It follows that by [1.29, Cor. 1], we have that ZEDC = ZDEF = ZEFA = ZDBF Similarly by [1.29, Cor. 1], we have that ZCED = ZEAF = ZFDB = ZDFB. Since BAFDE is a parallelogram, we have that ZEAF = ZEDF [1.34], or ZCED = ZEAF = ZEDF = ZDFB 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 339 Since BEFCD is a parallelogram, we have that EF = CD. Since BEFBD is a parallelogram, we have that EF = BD, or EF = CD = BD By [1.26], AAEF = ADEF = AECD = DBF \u25a1 [1.46] Exercises 1. Two squares have equal side-lengths if and only if the squares are equal in area. PROOF. Suppose we have two squares with equal side-length. Divide each square into two equal triangles by constructing a diagonal. The side-length is then the altitude of each triangle. By [1.34], each triangle is half of the area of the square. By [1.40, #1], each triangle is equal in area to its corresponding tri- angle in the other square. Together, these results show that all four triangles are equal in area, and therefore the squares are equal in area. Now suppose that we have two squares which are equal in area where the side-lengths are unequal. Divide each square into triangles by the above method. It follows that each side of a triangle from the larger square has a longer altitude than its corresponding side on the smaller triangle. It follows that the areas of the triangle are unequal. Thus, the proof. \u25a1 [1.47] Exercises 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 340 10. Each of the triangles AAGK and ABEF formed by joining adjacent corners of the squares is equal in area to the right triangle AABC. PROOF. Construct the polygons as in [1.47] and then construct segments AG and EF. We wish to show that AABC = AAGK = ABEF We will employ trigonometry to solve this exercise. First, we note that one right angle equals | radians. Consider AABC. Notice that if ZCAB = 7, then AABC = \u00a7 – 7 since ZACB = \u00a7 . It follows that ZKAG = 2tt – (| + I + ZCAB) = ir + 7 and that ZFBE = 2tt – (| + I + ZABC) = I – 7 We will employ the general form of the equation of the area of a triangle: Area = -xy \u2022 sin# where # is the interior angle to sides x and y. Area AAKG = AK \u25a0 AG \u25a0 sin(n + 7) = AK \u25a0 AG \u2022 sin( 7 ) 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 341 by the properties of the sin function. We also have that Area AABC = AB \u2022 AC \u2022 sin( 7 ) and so Similarly, and = AK \u2022 AG \u2022 sin( 7 ) AAKG = AABC Area ABE F = BE \u2022 BF \u2022 sin(7r + 7) = BC- AB-cos( 7 ) Area AABC = AB \u2022 BC \u2022 sin(f – 7) = AB \u25a0 BC \u25a0 cos(j) by the properties of the cosine function. Thus, the proof. \u25a1 Chapter 1 exercises 1. Any triangle is equal to a fourth part of the area which is formed by constructing through each vertex a line which is parallel to its opposite side. PROOF. Construct AABC; then construct DF \\ BC through point A, con- struct EF || AB through point C, and construct DE \\ AC through point B. We wish to show that the area of ADEF is equal to four times the area of AABC. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 342 FIGURE 1.0.35. Chapter 1 exercises, #1 Since DF \\ BC, ZADB = ZCBE. Since DE \\ AC, ZADB = ZCAF. Hence, ZADB = AC BE = ZCAF Similarly, we can show that ZCFA = ZBAD = ZECB By [1.32], it follows that ZDBA = ZBEC = ZACF Notice that BAFCB is a parallelogram. Hence, AF = BC. Since BDACB is also a parallelogram, BC = DA. By [1.26], AABD = ACEB = AFC A By [1.8], AABD = ACEB = AFC A = AABC Since ADEF = AABD \u00a9 ACEB \u00a9 AFC A \u00a9 AABC, the proof follows. \u25a1 4. The three medians of a triangle are concurrent. (Note: we are proving the existence of the centroid of a triangle.) PROOF. Construct AABC where CE is the median of AB and BF is the median of AC. Through the intersection of CE and BF, point G, construct the segment AGD where D is a point on BC. We wish to show that BD = DC. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 343 FIGURE 1.0.36. Chapter 1 exercises, #4 By [1.38, #1], notice that ABGE \u00a9 AEG A \u00a9 AAGF = AFGC \u00a9 ACGD \u00a9 ADGB and ACGD \u00a9 ADGB \u00a9 ABGE = AEG A \u00a9 AAGF \u00a9 AFGC It follows that ABGE \u00a9 AEG A \u00a9 AAGF \u00a9 AFGC = ACGD \u00a9 ADGB 2 \u2022 ABGE \u00a9 AEGA \u00a9 AAGF \u00a9 AFGC = ACGD \u00a9 ADGB \u00a9 ABGE 2 \u2022 ABGE \u00a9 AEGA \u00a9 AAGF \u00a9 AFGC = AEGA \u00a9 AAGF \u00a9 AFGC 2 \u2022 ABGE \u00a9 AFGC = AFGC 2 \u2022 ABGE = 2 \u2022 AFGC ABGE = AFGC By [1.38], we have that ABGE = AEGA and AFGC = AFGA, and so ABGE = AEGA = AFGC = AFGA Suppose i^, a point on \u00a3?D other than 5 or F>, is the midpoint of BC. Then BH = HC and by [1.38], ABGH = ACGH, or ABGH = ACGD \u00a9 AHGD ABGH \u00a9 AHGD = ACGD Also notice that by [1.38, #1], AABD = AACD AEGA \u00a9 ABGE \u00a9 ABGH \u00a9 AHGD = AFGA \u00a9 AFGC \u00a9 ACGD ABGH \u00a9 AHGD = ACGD 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 344 By the above, we have that ABGH Q AHGD = ABGH \u00a9 AHGD which can only hold if AHGD has no area, a contradiction. A similar result follows if H is a point on CD which is neither C nor D. Thus, BD = DC. \u25a1 COROLLARY. The three medians of a triangle divide the original triangle into six sub-triangles which are each equal in area. 8. Construct a triangle given the three medians. PROOF. Suppose we are given the medians of a triangle AE, BF, and CD. We wish to construct the triangle to which they belong. FIGURE 1.0.37. Chapter 1 exercise #8 Connect points A and B; connect A and C; finally, connect B and C. Since AE is a median, BEC forms a segment where BE \u2014 EC. Similar statements can be made for the remaining side, mutatis mutandis. Hence, the required triangle has been constructed. \u25a1 12. The shortest median of a triangle corresponds to the largest side. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 345 PROOF. Construct AABC where BC > AC > AB, CE is the median of AB, BF is the median of AC, and AD is the median of BC. We wish to show that AD < FB and AD BE (because BC > AB, BD = BC. and BE = BA) , and since the triangles share side BG, we must have that EG > DG. Similarly, we can show that EG > FG > DG. By [Ch. 6 Exercises. #124], we have that the medians of a triangle divide each other in the ratio of 2 : 1. Hence, EC > FB > AD. \u25a1 16. Inscribe a lozenge in a triangle having for an angle one angle of the triangle. PROOF. Construct AABC. Let ZDAB bisect ZCAB where D is a point on the side BC. Construct the ray AD. Construct ZADF and ZADE such that ZADF = ZDAB = LADE We claim that [HAEDF is the required lozenge. 1. SOLUTIONS: ANGLES, PARALLEL LINES, PARALLELOGRAMS 346 Consider ADFA. Since it is an isosceles triangle, DF = FA [1.6]. Also consider ADEA. Notice that by a similar argument, we have that AE = ED. Since ADFA and ADEA share side AD, by [1.26] we have that ADFA 9* ADEA. Hence, DF = FA = AE = ED By [Def. 1.29], BAEDF is a lozenge. Since AABC and BAEDF share ZBAC, the proof follows. \u25a1 CHAPTER 2 Solutions: Rectangles [2.4] Exercises 2. If from the vertical angle of a right triangle a perpendicular falls on the hypotenuse, its square equals the area of the rectangle contained by the segments of the hypotenuse. PROOF. Construct right triangle A ABC where ABAC is the vertical angle. Construct segment AD such that AD _L BC. We claim that AD 2 = DB \u25a0 DC. F > Figure 2.0.1. [2.4, #2] Construct rectangle [HDCHG where BD = DC Geometrically, we claim that AD 2 = BDCHG. By [1.47], we have that AD 2 + DC 2 = AC 2 AD 2 + DB 2 = AB 2 347 2. SOLUTIONS: RECTANGLES 348 as well as AB 2 + AC 2 = (DB + DC) 2 AB 2 + AC 2 = L>\u00a3 2 + 2-DB-DC + DC 2 Hence, AD 2 + DC 2 + AD 2 + L>\u00a3 2 = L>\u00a3 2 + 2 \u2022 L>\u00a3 \u2022 DC + DC 2 2 \u2022 AD 2 = 2-DB-DC AD 2 = DBDC \u25a1 [2.6] Exercises 7. Give a common statement which will include [2.5] and [2.6]. PROOF. Let AB be a line and consider the segment formed by points A and B. Locate the midpoint C of this segment. Choose a point D on the line AB such that D is neither A, B, nor C. We then have two cases: 1) D is between A and 5. By [2.5], we have that AD.DB + CL> 2 = C\u00a3 2 2) D is not between A and B. By [2.6], we have that AD.DB + C\u00a3 2 = CD 2 \u25a1 [2.11] Exercises 3. If AB is cut in “extreme and mean ratio” at C, prove that (a) AB 2 + \u00a3C 2 = 3AC 2 PROOF. By [2.11], x = \u2014 f (1 \u00b1 -s\/5). (We may ignore negative results since our context is the side-length of planar figures.) Since AB = a, BC = a \u2014 x, 2. SOLUTIONS: RECTANGLES 349 and AC = x, we have: AB 2 + BC 2 a 2 + (a – x) 2 ! + a 2 – 2ax + x 2 2a 2 – 2ax a 2 \u2014 ax a 2 + ^(l + V5) ^(3 + >\/5) 3.4C 2 3x 2 3x 2 2x 2 x 2 4(3- 4(3- n\/5) V5) \u25a1 Chapter 2 exercises 15. Any rectangle is equal in area to half the rectangle contained by the diagonals of squares constructed on its adjacent sides. 2. SOLUTIONS: RECTANGLES 350 Let AB = x and BC = y. By [1.47], it follows that GB = xV2 and BC yV2. Notice that GB 2 \u25a0 BF 2 = xyy\/l = xy = AB-BC \u25a1 CHAPTER 3 Solutions: Circles [3.3] Exercises 3. Prove [3.3, Cor. 4]: The line joining the centers of two intersecting circles bisects their common chord perpendicularly. PROOF. Construct the circles from [1.1], and then construct segment CF. We wish to show that AB bisects CF and that AB _L CF. Figure 3.0.1. [3.3, #3] The proof follows immediately from the proof of [1.1, #2]. The details are left as an exercise to the reader. \u25a1 [3.5] Exercises 2. Two circles cannot have three points in common without coinciding. PROOF. Suppose that two circles (oEDF, oEBF) have three points in com- mon (E, F, and G) and do not coincide. By [3.3, Cor. 4], the line joining the centers of two intersecting circles (AC) bisects their common chord perpen- dicularly. Hence, AC bisects EF; similarly, AC bisects EG. But EG can be constructed to that EG and AC do not intersect, a contradiction. 351 3. SOLUTIONS: CIRCLES 352 Figure 3.0.2. [3.5, #2] Hence, oEDF and oEBF coincide. \u25a1 [3.13] Exercises 3. Suppose two circles touch externally. If through the point of intersec- tion any secant is constructed cutting the circles again at two points, the radii constructed to these points are parallel. PROOF. Suppose oGBD and oEFB touch at point B. By [3.13], these cir- cles touch only at B. Construct secant DBE. We claim that AD \\ CE. Figure 3.0.3. [3.13, #3] 3. SOLUTIONS: CIRCLES 353 Connect AC. By [3.12], AC intersects B. Consider AABD and AC BE. Notice that ZABD = AC BE by [1.15]. Also, since each triangle is isosceles, ZADB = ZABD and ZCEB = AC BE. Hence, ZADB = ZCEB By [1.29, Cor. 1], AD \\ CE. \u25a1 COROLLARY. 1. If two circles touch externally and through the point of intersection any secant is constructed cutting the circles again at two points, the diameters constructed to these points are parallel. 4. Suppose two circles touch externally. If two diameters in these circles are parallel, the line from the point of intersection to the endpoint of one di- ameter passes through the endpoint of the other. PROOF. Suppose oDBE and oFBG touch at point B. By [3.13], these cir- cles touch only at B. Construct diameters DE and FG such that DE \\ EG. We claim that the line DB intersects G. (The case for the remaining endpoints follows mutatis mutandis.) Figure 3.0.4. [3.13, #4] Suppose that DB does not intersect G. Extend FG to a line, and suppose that DB intersects the line FG at H. Construct the segment AC. By [1.15], AABD = ACBH. By [1.29, Cor. 1], ZADB = ZCHB. It follows that AABD and ACBH are equiangular. How- ever, AABD is an isosceles triangle and ACBH is not since CH > CG; hence, the triangles are not equiangular, a contradiction. A similar contradiction is 3. SOLUTIONS: CIRCLES 354 obtained if DB intersects the line FG at any point other than G, mutatis mu- tandis. The proof follows. \u25a1 [3.16] Exercises 1. If two circles are concentric, all chords of the greater circle which touch the lesser circle are equal in length. PROOF. Construct oBFG and oDCE each with center A. On oDCE, con- struct chord DE such that DE touches oBFG at B. Also on oDCE, construct chord HJ such that HJ touches oBFG at G. We claim that DE = HJ. Figure 3.0.5. [3.16, #1] Notice that AG = AB since each are radii of oBFG. By [3.16], DE and HJ have no other points of intersection with oBFG. Hence, DE and HJ are equal distance from the center of oBFG and also of oDCE. By [3.14], DE = HJ. \u25a1 [3.17] Exercises 3. If a parallelogram is circumscribed to a circle, it must be a lozenge, and its diagonals intersect at the center. 3. SOLUTIONS: CIRCLES 355 PROOF. Construct circle oGHI with center E and circumscribe parallelo- gram BBC DA to oGHI where BBCDA touches oGHI at F, G, H, and I. We claim that BBC DA is a lozenge and that its diagonals AC and BC intersect E. Figure 3.0.6. [3.17, #3] Since BBC DA is a parallelogram, EA = EC and EB = ED. Also, BC = AD and BA = CD. By [1.8], AEAD 9* AECB and A\u00a3\u00a3A ^ A\u00a3DC. Now construct radii EH, EI. Since oGHI touches BBCDA at H and at \/, we must have and that ZEHD = ZEID, since each are tangents to oGHI and so are right angles; therefore, AEHD and AEID are right triangles. Since AEHD and AEID share side ED and EH = EI (since both are radii oioGHI), by [1.26, #6] we have that A\u00a3#L> ^ A\u00a3LD. Similarly, A\u00a3#A ^ AEIC, and so it follows that AEAD ^ AECD. It follows that AD = CI} = BA = \u00a3C and so BBC DA is a lozenge. Suppose E is not the center of oGHI. But I? is a point within a circle from which three or more equal segments can be constructed to the circumference, a contradiction [3.9]. Hence, the diagonals of BBC DA intersect E. B [3.30] Exercises 3. SOLUTIONS: CIRCLES 356 1. Suppose that ABCD is a semicircle whose diameter is AD and that the chord BC when extended meets AE (where AE is the extension of AD). Prove that if CE is equal in length to the radius, the arc AB = 3 \u2022 CD. PROOF. Suppose ABCD is a semicircle where BC is a chord such that if BC and AD are extended, they intersect at E, and that CE = DF. We wish to show that arc AB = 3 \u2022 CD. Figure 3.0.7. [3.30, #1] Connect CD, CF, CB, FB, and AB. Notice that CE = DF = CF = BF = AF BY [1.5], we have that ZCEF = ZCFE, ZFCB = ZFBC, and ZFBA = ZFAB. We will solve this problem using linear algebra. Let ZCEF = a ZECF = b ZFCB = c ZCFB = d ZFBA = e ZBFA = f Then we have that b + c = 180 2a + b = 180 2c + d = 180 2e + \/ = 180 a + d + \/ = 180 a + c + 2e = 180 3. SOLUTIONS: CIRCLES 357 In matrix form, this is 0 1 1 0 0 0 180 2 1 0 0 0 0 180 0 0 2 1 0 0 180 0 0 0 0 2 1 180 1 0 0 1 0 1 180 1 0 1 0 2 0 180 This matrix in reduced row echelon form is: Or, 1 0 0 0 0 0 20 0 1 0 0 0 0 140 0 0 1 0 0 0 40 0 0 0 1 0 0 100 0 0 0 0 1 0 60 0 0 0 0 0 1 60 ZCEF 20\u00b0 ZECF 140\u00b0 ZFCB 40\u00b0 ZCFB 100\u00b0 ZFBA 60\u00b0 ZBFA 60\u00b0 Since ZCFD = ZCFE = ZCEF, ZBFA = 3 \u2022 ZCFD. By [7.29, Cor. 1], it follows that AB = 3 \u2022 CD. \u25a1 CHAPTER 4 Solutions: Inscription & Circumscription [4.4] Exercises i Figure 4.0.1. [4.4, #1] 1. In [4.4]: if the points O, C are joined, prove that the angle ZACB is bisected. Hence, we prove the existence of the incenter of a triangle. PROOF. Consider AOFC and AODC. By the proof of [4.4], we have that ZOFC = ZODC since each are right; OF = OD since each are radii of oDEF; and each shares side OC. By [1.26, #6], AOFC ^ AODC, and so ZOCF = ZOCD. Hence, ZACB is bisected, and O is the incenter of AABC. \u25a1 [4.5] Exercises 1. The three altitudes of a triangle (AABC) are concurrent. (This proves the existence of the orthocenter of a triangle.) PROOF. Euler’s proof: consider a triangle AABC with circumcenter O and centroid G (i.e., the point of intersection of the medians of the triangle). 358 4. SOLUTIONS: INSCRIPTION & CIRCUMSCRIPTION 359 Figure 4.0.2. [4.5, #1] Let A! be the midpoint of BC. Let H be the point such that G is between H and O and EG = 2 \u2022 GO. Then the triangles AAGH, AA’GO are similar by side-angle-side similarity. It follows that AH \\ OA’ and therefore AH _L \u00a3?C; i.e., it is the altitude from A. Similarly, BH, CH, are the altitudes from B, C. Hence all the altitudes pass through ilQ \u25a1 http : \/\/www . artof problemsolving . com\/Wiki\/ index . php\/Orthocenter
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