Geometry for College Students

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Geomet forColege Sdents

I. Martin Isaacs

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GEOMETY


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he rooksoe Series in dvanced Mathematicsau J Say Jr Editor

Probabiliy The Science of Uncertainywith Applications to InvestmentsInsurance and Enginee ringMichae eanniversity of Weste Ontario001 ISN: 05343031

The Mathematics of FinanceModeling & HedgingJoseph Stampiictor Goodmanniversity of Indiana oomington001 ISN: 053437779

Geomet for College Students

I Martin Isaacsniversity of Wisconsin Madison001 ISN: 0534351794

A Course in Approximation TheoWard heneyhe niversity of exas ustinWi ightnivesity of eicester Engand000 ISN: 0534349

Introduction to Analysis Fth EditionEdward GaughanNew Mexico State niversity1998 ISN: 0534351778

Numerical Analysis Second Editionavid ncaidWard heneyhe niversity of exas ustin199 ISN: 053433895

Advanced Calculus A Course inMathematical Analysisatick M Fipaicknivesity of Maryand199 ISN: 053491

Algebra A Graduate CourseI Matin Isaacsniversity of Wisconsin Madison

1994 ISN: 05341900

Fourier Analysis and Its ApplicationsGerad Foandniversity of Washington199 ISN: 0534170943


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saacs, Martin, ( date)Geomety for college students Martin saacs

p cmncludes indexSBN 0-534-35179-4 (text)1. Geometry Title

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Preface

his book is primariy intended for coege mathematics students who enjoyed highschoo geometry and who wish to ea more about the amaing properties of inescircles trianges and other geometric gures I n particuar I hope that those who arepreparing to become highschoo mathematics teachers wi nd inspiration here to hepthem share their enthusiasm and enjoyment of geometry with their own future students

ut why shoud anyone study geometry? One reason of course is that geometry andits descendant trigonometry are essentia toos in engineering architecture navigationand other discipines hese practica appications however surey do not expain whyit is that for centuries geometry has been taught to amost every student and why at eastuntil recenty a person who knew no geometry was not considered to be propery educated I think there are at east two reasons more important than usefuness that expainwhy geometry has been and shoud continue to be a part of the schoo curricuum

Since Eucid some 300 years ago geometry has been taught as a deductive sciencewith theorems and proofs s a consequence, generations of geometry students haveeaed how to draw vaid concusions from hypotheses and how to detect and avoid

invaid reasoning In other words by studying geometry students can e how to thinkOf course there are other subjects that coud aso be used to teach deductive reasoning but geometry is especiay eective because it seems to have the perfect baance ofdepth and concretenes s Many of the theorems that we prove in geometry are dee in thesense that they assert something nonobvious and sometimes even surprising hey areconcrete because students can easiy draw the appropriate trianges circes or whatever,and they can see that what is aeged to happen actuay does appear to happen

Geometry is aso beautifu and some of its theorems are so amaing as to seemamost mracuous In fact much the same comment coud be made about most areasof mathematics but geometry is unique in that its miraces are visua so they can

readiy be appreciated even by the uninitiated Surey one does not need a great deaof mathematica sophistication to marve at the fact that if a ine is drawn through eachvertex of any triange and if each of these ines is perpendicuar to the opposite side ofthe triange then these three ines a go through a common point One coud argue thatthe fundamenta theorem of cacuus for exampe is equay amaing and beautifubut unfortunatey it is not poss ibe to appreciate it without rst studying cacuus utthe aesthetic vaue of geometry extends beyond the strikng statements of its theorems


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PREACE

Many of the proofs have their own subte and eegant beauty which unfortunatey isa itte harder to perceive. Nevetheess we expect that most readers of this book wiea to enoy the beauty of the proofs as we as that of the theorems .

In shot we shoud continue to study and teach geometry because it is a highyattractive subect and because we can ea from it something about deductive reasoningand the nature of mathematica proof. It seems cear therefoe that as in the past studentsshoud continue to see theorems being proved in their geometry cass they should betaught how to understand proofs and perhaps even more impotant they shoud leahow to invent and write proofs.

I have seected for this book some of the more spectacuar theorems in pane geometry and I have presented ustications of these facts using a vaiety of dierenttechniques of proof. Mosty ignored however are the kind of unsurprising theoremwhich whie required by mode standards of mathematica rigor can seem ratherpointess to students It requires considerabe sophistication to appreciate why anyonewoud want to prove a fact that seems competey obvious . Even professiona mathematcians, most of whom do understand the signicance of these resuts often nd theirforma axiomatic proofs somewhat du earning proofs shoud not be an unewardingchore instead we expect that students wi demand proofs because the assertions beingestabished are otherwise so incredibe.

is book was written as a text for oege Geometry which is a course thatI have taught severa times at the niversit of Wisconsin Madison. The coursesprincipa audience consi sts of sophomore and unior undergraduate math maors whoare speciaiing in secondary education. For them the course is required but there isaso a substantia minority who take the course eectivey Some of these students simpywant to earn geometry whie others take oege Geometry because they nd it tobe a more gente and accessibe introduction to mathematica proof than a course inabstract agebra or advanced cacuus . It is my beief that for some students the study

of geometry is an exceent preparation for these more difcut abstract courses .I have been dissatised with the avaiabe texts that might be used for suh a

course . Many incude a umber of interesting topics within a arge samping of assortedgeometric materia. ut they do not put the focus where I think it beongs: on thereay pretty theorems and their proofs that in my opinion shoud be at the heat of ageometry course for coege students. so they often devote much more space than Ithink is appopriate to fomaism and axiomatics Most students probaby do not ndthis especiay exciting and I share that opinion.

here aso exist some wonderfu books that are ed with spectacuar theoremsand eegant proofs but none of these seems quite suitabe as a text for this geometry

course either I have found that most students who register for oege Geometry claimto remember very itte from their one previous exposure to geometry in high schoo.For the sake of this maority some review and accimatiation are necessary. textis needed therefore that starts from a point cose to the beginning and introduces thenotion of proof genty and then gets to the good stu" as quicky as pos sibe . ecauseI coud not nd a book that covered the right materia at the right pace and that statedfrom the right pace I taught the course many times without a text.


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PREACE

Most of my students seemed to enjoy the course and many of them became veryexcited about geometry. In this book which is an expansion of my course lecture notesI have tried to reproduce as closely as pos sible the experience of the classroom and soI hope that my readers will also nd that geometry is an enjoyable and exciting subjectMost of all I hope that those of my students and readers who are or will be teachinghighschool mathematics wil l convey some of that excitement to their own students

I am grateful to the following reviewers for their helpful comments :Fred Flene, Notheaste llinois Univesity

Aa Hoe Uivesity of Caifoia Ivne

Kathyn Lenz, Uivesity of Minesota, Duuth

David Poole, Tent Univesity

Ron Soomo, Ohio State Univesity

Lay Sowde, San Diego State Univesity

Alex Turull Univesity of Florida, Gansville

Jeane Wald Michigan State Uivesity

I Martin Isaacs


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Contents

CHPER 1

Introduction and poogy 1 ongruent rianges 51 nges and arae ines 1 11 araeograms 14E rea 18

1 F irces and rcs 23G oygons in irces 341 Simiarity 39

CHPER 2

52 he ircumcirce 502 he entroid 5

2 he Euer ine, Orthocenter and Nineoint irce 02 omputations 72E he Incirce 7 32F Exscribed irces 802G Moreys heorem 822 Optimization in rianges 85

CHPER 3

43 Simson ines 943 he uttery heorem 1053 ross atios 103 he adica xis 1 19


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CONTENTS

CHPER 4

v' v 54 eva s heorem 1 254 Interior and Exterior evians 1 3 14 eva s heorem and nges 1 3

4 Meneaus heorem 14

CHPER 5

V M f f 565 ectors 155 ectors and Geometry 1585 ot roducts 1 35 heckerboards 1

5E it of rigonometry 1705F inear Operators 172

CHPER 6

ues of the Game 1 82 econstructing rianges 1 87

angents 191 hree ard robems 1 9E onstructibe Numbers 203F hanging the ues 208

6


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GEOMETRYfor

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C H P E R O N E

Te Basics

A ntroduction and Apology

Since the appearance about 2300 years ago of Eucids book The Elements, it has beentradtiona to study geometry as a deductive discipine, using the socaed axiomaticmethod. Ideay this means that the geometer (or geometry student) starts with a fewassumed facts caed axioms or postuates and then systematicay and carefuy derivesthe entire content of the subject using nothing but pure ogic . he axiomatic methodrequires that this content, which for geometry is a coection of facts about triangescirces and other gures shoud be presented as a sequence of ever deeper theoremseach rigorousy proved using the axioms and earier theorems .

Geometry has continued to deveop in the centuries since Eucid . Mathematicansincuding many taented amateurs have discovered a weath of beautifu, surprising and

even spectacuar properties of geometric gures and these together with the geometryof The Elements, constitute what is today caed Eucidean geometry. mazing as manyof these asserted facts may be we can be condent that they are correct because theyare proved they are not merey tested by experiment and conrmed by measurementand observation. Furthermore we do not have to rey on the authority of Eucid or hissuccessors, because with a itte practice and preparation, we can read and undertandthese proofs ourseves. With uck we might nd simper and more eegant proofs forsome of the known theorems of Eucidean geometry. It is even possibe that we mightdiscover new geometric facts that have never been seen before and invent our own proofsfor them.

he deveopment of geometry has resuted not ony in the discovery of amazingfacts, but aso in the invention of new and powerfu techniques of proof and these tooare considered part of Eucidean geometry. escartes invention of coordinate geometryis an exampe of this. third thread in the history of geometry, especiay in modetimes has been an investigation into the foundations of the subject . It has been askedfor instance whether or not a of Eucid s axioms and postuates are reay vaid andwhether or not we reay need to assume them a. ut we say amost nothing aboutthese foundationa issues in this book. Instead, we concentrate on the two more cassicathemes of geometry: facts and proofs. We oer the reader a seection of some of the


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2 CHAPTER THE BASCS

more striking facts of Eucidean geometry and we present enough proof machinery toestabish them.

In the centuries since Eucid the accepted standards of mathematica rigor havechanged and the sufciency of proofs in the stye of Eucid has been chalenged. Oneobjection for example is that Eucid reied too heaviy on diagrams and that he andthe other cassica geometers did not aways prove facts they considered to be obviousfrom the gures. In this book we foow the tradition of Eucid and of most of hissucces sors and so we are wiing to use some of the information contained in carefuydrawn gures . ut we must not rey on diagrams for certain other types of infoation,and unfortunatey it is difcult to be precise about exacty what we are wiling to readoff from a diagram and what requires a proof. We are condent that readers wil catch onto this fairly quicky however and we apoogie for the ambiguity. erhaps an examplewi hep carify the situation.

In Figure 1 . 1 ange bisectors A B , and CZ have been drawn in ABC hreefacts that seem cear in the gure are:

A

B

---

----

C

Figure 11

1 B < C In other words line segment B is shorter than ine segment C

2. he point , where the bisector of LA meets ine BC, ies on the ine segment BC3 he three ange bisectors are concurrent.

efore we discuss these observations et us be clear about the meanings of thetechnica words and notation. We assume that the nouns point, line, angle, triangle,and bisector are famiiar to readers of this book. wo ines, uness they happen to beparae aways meet at a point but if three or more ines a go through a common point,then something unexpected is happening and we say that the ines are concurrnt heremaining undened technica word in the precedng ist of facts is segment linemnt is that part of a ine that ies between two given points on the ine. Observe thatin assertion ( 1 ) we are using the notation B in two dierent ways : In the inequality,B represents the length of the ine segment, which is a number and then ater B isused to name the segment itsef. In addition the notation B is often used to representthe entire ine containing the points B and , and not just the segment they determine.ut we sha always provide enough infoation so that it is cear from context whichof the three pos sible meanings is intended. For exampe in inequaities or equations itis obviousy the numerica interpretation that is wanted.

Each of ( 1 ) (2) and (3) is true but we need to distinguish among three dierenttypes of truth here. First note that the fact in (1) that B < C is an accident. his


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A NTRODUCTON AND APOLOGY

inequality happens to be true in this gure, but it is not an instance of some general oruniversal truth Even in a particular case, this sort of information can be unreliable sinceit depends on the accuracy of the diagram and the care with which measurements aremade It is never considered valid to read o an equality from a diagram and it is rarethat one is justied in reading off an inequality

Fact (2) that point lies between points B and on line B , is not accidentalIndeed, it seems completely obvious that the bisector of each angle of an arbirarytriangle must always intersect the opposite side of the triangle In other words, thebisector intersects the line segment determined by the other two vertices of the triangle Note that although this is true about angle bisectors , it can fail for other important linesassociated with a triangle For example the attud drawn from A, which is the linethrough A perpendicular to line B , may not meet the segment B lthough thisfailure (accidentally) does not happen for AB of Figure 1 1 , it certainly can happenfor other triangles he obvious" fact that angle bisectors of triangles meet the oppositesides is the sort of information that Euclid was, and we are willing to read off fromdiagrams

lthough it is a fact that for every triangle, the three angle bisectors are concurrent,as in assertion (3) we follow Euclid in that we do not cons ider this to be obvious heexperimental evidence of even an accurate computerdrawn diagram (or of many suchdiagrams) is not sufcient for us to accept this as a universal truth we require a proofeaders will probably have seen a proof of this theorem in highschool geometry, and itis also proved here in hapter 2

hus there is an inherent ambiguity about which information can be reliably established from diagrams and which cannot ecause of this ambiguity, modern standards ofmathematical rigor require that there must be no reliance whatsoever on gures Geometry without diagrams was made possible by avid ilbert (1 821943) who, building onthe work of his predecessors constructed an appropriate set of precisely stated axiomsfrom which he was able to prove everything formally and without diagrams Euclid sideal that geometry should be a purely deductive enterprise was thus nally realized byilbert at the turn of the 20th century

In particular ilberts axioms allowed him to prove our observation (2), that anangle bis ector of a triangle always meets the opposite side of the triangle ut it is farfrom a triviality to prove rigorously, and without relying on a diagram, that the bisectorof B A meets line B at a point that lies between B and Furthermore, therewould be no hope of proving such a thing without having a precise denition of theword between, which ilbert was able to provide

he achievements of ilbert and other researchers into the foundations of geometry

were substantial and signicant, but it is my opinion that, by far, the most interestingtheorems of geometry are those that provide surprises We feel, for example, that thefact that the three angle bisectors of a triangle are always concurrent is much moreexciing than is the fact that the angle bisectors meet the opposite sides of the triangleFor his reason we have chosen to present in this book as many unexpected facts andsurprising theorems as space allows Since we want to get to these quickly we must omitilbert s careful and rigorous treatment of the foundations of geometry, and instead, wewill follow tradition and rely on diagrams without specifying exactly the extent of our


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4 CHAPTER THE BASCS

reiance. so since the reader of this book wi have studied some geometry in highschoo we wi not start our presentation at the very beginning of the subject.

We have aready apoogized for the ambiguity about how much information we areaowed to obtain from diagrams. Some apoogy is aso appropriate concerning the issueof how much highschoo geometry we are assuming. Students may fairy compainwhen they are doing homework exercises that it is uncear which facts they must proveand which they can merey quote as remembered from schoo . We do not attempt to givea compete ist of assumed resuts but we sha show by exampe the eve of proof thatwe expect and we sha devote much of the rest of this chapter to a review of some ofthe essentia facts denitions and theorems from highschoo geometry.

efore proceeding with our review of highschoo geometry we discuss brieysome of the issues in the foundations of geometry to which we referred earier Wementioned that Eucid caed his unproved assumptions axioms and postuates. hedistinction, which is not considered signicant today is that Eucid s axioms conceedgenera ogica reasoning, whie his postuates were more specicay geometric. Forexampe one of Eucid s axioms is things equa to the same thing are equa to eachother" whereas his parae postuate essentiay asserts that given a ine and a pointnot on that ine there exists one and ony one ine through the given point parae tothe given ine." ctuay Eucid s parae postuate is not stated in precisey this way itappears in a somewhat more compicated but ogicay equivaent formuation.

Over the centuries, Euclid s parae postuate has engendered a great dea of interestand controversy. Somehow the existence and uniqueness of a ine parae to a given inethrough a given point seemed ess obvious than the facts asserted by the other postuates.Geometers fet uncomfortabe assung the parae postuate and they attempted insteadto prove it they tried to deduce it from the rest of Eucid s axioms and postuates eforewe discuss these attempts we shoud stress that the parae postuate makes two separateassertions each of which woud have to be proved. It woud be necessary to show that

a parae ine (through the given point) exists, and one woud aso need to prove that itis unique

ow might a proof of the parae postuate proceed? One coud assume that it isfase and then try to derive some contradictory concusions s sume for exampe thatthere is some ine AB and some point P not on AB, and there is no ine parae to ABthrough P If by means of this assumption one coud deduce the existence of a triangeZ for which Z and aso < Z, then this contradiction woud proveat east the existence part of the parae postuate. What actuay happened when thiswas tried was that apparent contradictions were derived and the existence of gures thatseem impossibe was proved. For exampe the assumption that no ine parae to AB

goes through point P yieds a triange Z that has three right anges . Surey this isimpossibe" we might say but how can we prove this impossibiity? We know from highschoo that the three anges of a triange sum to 1 80 and an easy experiment conrmsthis fact (ear o the three coers of a paper triange and ine them up to see thatthe three anges tota a straight ange. ) It may seem that by proving the existence of atriange with three rght anges, we have the desired contradiction but this is wrong. hehighschoo proof that L + L + L Z 1 80 utimatey reies on the parae postuatewhich we are temporariy refusing to assume. so, the experiment with paper triangesis certainy not a mathematica proof.


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B CONGRUENT TRANGLES

fter repeated attempts to obtain contradictions from the denia of either the existence or the uniqueness part of the parae postuate it was reaized by J oyaiN. obachevski and Gauss in the 19th century that athough such denias yiedseemingy ridicuous situations such as trianges with three right anges no proof of acontradiction was pos sibe . In fact it was proved that no such proof is possibe. In otherwords one coud buid a perfecty consistent deductive geometry by repacing Eucid sparae postuate with either one of two ateative new postuates. One of these deniesthe existence of a parae to some ine through some point and the other asserts theexistence of at east two such paraes . Each of the two types of geometry that arise inthis way is said to be nonEucidean and each has its own set of proved theorems. hegeometry where no parale exists i s caed eiptic geometry, and when more than oneparae to a ine goes through a point we have hyperboic geometry. he deductions ofeach of the two types of nonEucidean geometry contradict each other and they asocontradict the theorems of cassica Eucidean geometry but each of the three avors ofgeometry appears to be inteay consistent and from a mathematician s point of viewthey are equay vaid. ctuay it is known that if Eucidean geometry is internayconsistent then the two nonEucidean geometries are aso consistent but no formaproof of the consistency of Eucidean geometry has been found.

he question of which if any of the three avors of geometry that we have beendiscussing describes the rea word is interesting, but it is not reay a mathematicaquestion it beongs in the ream of physics . he itte experiment with the paper triangecertainy suggests that we ive in a Eucidean universe but this has not been rmyestabished on a arge scae. Indeed modern theories of the structure of the cosmosincuding Einstein s theory of genera reativity suggest that none of the three geometriesprovides an entirey accurate description of the universe in which we actuay ive.Neverthees s Eucidean geometry provides a very good approximation to reaity on ahuman scae and so it is usefu for practica purposes such as engineering navigation,and architecture.

IB Congruent iangles

We devote the rest of this chapter to some of the basic facts and techniques of Eucideangeometry much of which wi be a review for most readers. eca that two geometricgures are conrunt if informay speaking they have the same size and shape.Somewhat more precisey two gures are congruent if one can be subected to a rigidmotion so as to make it coincide with the other. y a rigid motion we mean a transationor shift a rotation in the pane or a reection in a ine . he atter can be viewed informayas ifting the gure from the pane ipping it over and pacing it back in the pane

In Figure 1.2, for exampe the three trianges are congruent athough to makeZ coincide with either of the other two trianges it is necessary to reect it or ip itover. We write ABC RST to report that the rst two trianges in Figure 1 arecongruent but note that there is more to this notation than may at rst be apparent. heony way that these two trianges can be made to coincide is for point R to coincide withpoint A, for S to coincide with B, and for T to coincide with C We say that A and R Band S, and C and T are corrpondin point of these two congruent trianges . he ony


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6 CHAPTER THE BASCS

zR

s

T y

Figure 1.2

correct ways to report the congruence is to ist corresponding points in correspondingposi tions It is correct therefore to write ABC RST or BAC SRT butit is wrong to write ABC S RT his atter assertion is not true because there isno way that these two trianges in Figure 1 2 can be made to coincide with A and S, B,and R, and C and T being corresponding points

Since RST Z, it is cear that corresponding sides of these trianges haveequa ength and that corresponding anges have equa measure (contain equa numbers

of degrees or radians) We can thus write for exampe RS and LSRT L ZNote that in this context, the notation L S RT refers to the measure of the ange in someconvenient units such as degrees or radians In other situations however we may writeS R T to refer to the ange itsef his is entirey anaogous to the fact that R S can refereither to a ine segment or to its ength in centimeters inches mies or whatever Insome geometry books the notation mLSRT is used to denote the measure of LSRT

Whenever we know that two trianges are congruent we can deduce six equaitiesthree of engths and three of measures of anges It is aso reasonaby ovious that giventwo trianges , if a six equaities hod, then the trianges can be made to t" one ontop of the other and they are congruent s readers of this book are surey aware it is

not necessary to know a six equaities to concude that two trianges ae congruent Ifwe ow, for exampe that the three sides of one triange equa respectivey, the threecorresponding sides of the other triange we can safey deduce that the trianges arecongruent If we know, for exampe that AB RS, AC RT, and BC ST, we canconcude that ABC RST In a proof where each assertion must be justied wesay that these two trianges are congruent by SSS" he abbreviation SSS which standsfor sideside side" refers to the theorem that says that if the three sides of a triangeare equa in ength to the three corresponding sides of some other triange then the twotrianges must be congruent

Other vaid criteria that can be used to prove that two trianges are congruent

are SS S and S hese of course are abbreviations for sideangeside"angesideange" and sideangeange" respectivey In the expectation that theseare entirey famiiar to readers of this book we iustrate ony one of them with anexampe In Figure 12 if we somehow know that ST Z and that L S L andLR L, we can write in a proof We concude by S that SRT Z.(Notice that we wrote as a short form for LRST, which is the fu name of this angehis is acceptabe when it cannot resut in ambiguity)

What is the ogica status of the four congruence criteria SSS SS S, and S?For each of these the fact that the criterion is sufcient to guarantee the congruence of


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B CONGRUENT TRANGLES 7

two triagles is actually a theorem, proved by Euclid from his postlates These fourtheorems are amog the basic results that we are acceptig as kow to be valid ad thatwe are willig to use without providig proofs I fact, however, it is ot hard to provethe sufciecy of some of these criteria if we are willig to accept some of the otherss a example, ad as the rst proof that we actually preset i this book, let us deducethe sufciecy of the SSS criterio, with the uderstadig that we may freely use ayof the other three triaglecogruece coditios

(1.1) PROBLEM. ssume that i ABC ad R, we kow that AB R,AC R , ad BC rove that ABC R without usig the SSScogruece critero

To do this , we shall appeal to aother theorem that readers surely remember fromthei previous study of geometry: The base agles of a isosceles triagle are equalecall that a triagle U V W is ioc if two of its sides have equal legths IFigure 1 . 3 for example, the triagle is isosceles because UV UW. The third side V Wis called the ba of the triagle, whether or ot it actually occurs at the bottom of thediagram The ba an of a isosceles triagle are the two agles at the eds of thebase , ad the theorem asserts that they are ecessarily equal I Figure 1 . 3 therefore,we have U V W U W V. We metio that this baseagles theorem for isoscelestriagles is used so ofte that it is give a ame: the pon ainorum This ati phrasemeas bridge of asses " pparetly, the theorem has acquired this ame partly becausethe diagram i The Elements used i its proof vaguely resembles a bridge

Figre 13

oution to Probm 1.1. y reamig the poits , if ecessary, we ca assume thatAC is the logest side of ABC, ad cosequetly, R is the logest side ofR Sice we are give that AC R, we ca move R , ippig it over,if ecessary, so that poits A ad R coicide, as do poits C ad , ad so thatpoits B ad lie o opposite sides of lie A C

Now draw lie segmet B What must result is a situatio resemblig the left

diagram i Figure 14. It is ot possible for B to fail to meet A C, as i the ightdiagram of Figure 1 .4 or for B to go through oe of the poits A or C becausethat would require oe of BC or BA to be loger tha AC (We are shamelesslyrelyig o the diagram to see this ) We ca thus as sume that we are i the situatioof the left gure

Sice A B R , we see that A B is isosceles with base B , ad hece bythe pos asiorum, we deduce that y , where, as idicated i the diagram, weare writig ad y to represet the measures of AB ad RB, respectivelySimlarly, by a secod applicatio of the pos asiorum, we obtai u v ad it


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8 CHAPTER THE BASCS

A C

-

R T T

s s

Figre 1.4

follows by additio that x + u y + v I other words, LABC R Sicewe already kow that AB R ad BC , we ca coclude by SS thatABC R, as required

Observe that we have writte this proof i a coversatioal sty e, usig complete

Eglish seteces grouped ito logical paragraphs We eded by establishig what weset out to prove, ad we clearly marked the ed of the proof (It has become customaryfor a box to replace the oldfashioed E as a edofproof marker) This style, withmior variatios, has become the accepted model for what a proof should look likethroughout most areas of mathematics , ad we follow it cosistetly i this book Weexpect studets to do the homework exercises with proofs writte i the same style,usig complete seteces We do ot recommed the twocolum proof format that isofte required i highschool geometry classes

Note that the secod setece of the last paragraph of the proof begis with theword similarly This word ca be a powerful tool for simplifyig ad shorteig proofs

ad makig them more itelligible ike most powerful tools, however, this oe ca bedagerous if use improperly We ecourage studets to use the word similarly to avoiduecessary repetitio i their proofs, but to use it carefully

lthough we expect that most readers of this book remember how to prove thepos as iom, we preset more tha oe proof here to illustrate a few poits ad toprovide further models of proofwritig style Our rst proof also yields some additioaliformatio about isosceles triagles We remid the reader that a mdian of a triagleis the lie segmet joiig a vertex to the mdpoit of the opposite side

THEOREM et ABC be isosceles with base BC Then L B LC Also the

median from vertex A the bisector of L A and the altitude from vertex A are all thesame line

Proof. I Figure 1 . 5 we have draw the bisector A of LA , ad thus LBALCA y hypothesis, we kow that AB AC ad of course, A A ThusBA CA by SS It follows that L B LC sice these are correspodigagles i the coget triagles

We also eed to show that the agle bisector A is a media ad that it is aaltitude too To see that it is a media, it sufces to check that is the midpoit


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of segmet B C, ad this is true becauseB X X C sice these are correspodig sides of our cogruet triagles Fially, to prove that AX is also a altitude,we must show that AX is perpedicularto BC I other words, we eed to establish that B X A 90 . From the cogruettriagles, we kow that the correspodigagles B X A ad C X A are equal , adthus BXA BXC 90 sice thestraight agle BX C 180.

B CONGRUENT TRANGLES 9

A

B C

Figre 1.5

To prove that the agle bisector, media, ad altitude from vertex A are all the same,we started by drawig oe of these lies (the bisector) ad showig that it was aso amedia ad a altitude Sice there is oly oe media ad oe altitude from A, weko that the bisector is the media ad the altitude

What would have happeed if we had started by drawig the media from A isteadof the bisector of A? This is , after all, the same lie I this situatio, we could deducethat B AX CAX by SSS It would the follow that BAX CAX We woulddeduce that AX is the agle bisector, ad everythig would proceed as before Thisappoach would have bee less satisfactory, because it makes the pos asiorum depedo the SSS cogruece criterio I roblem 1 . 1 , however, we proved the validity of theSSS criterio usig the pos asiorum, ad this would be a example of ivalid circularreasoig

What if we had started by drawig altitude AX? We would the kow that BXA

C X A sice both of these are right agles Wealso kow that AB AC ad AX AX tthis poit, we mght be tempted to coclude thatBAX CAX by SS, but we would resistthat temptatio, of course, because SS is ot avalid cogruece criterio To see why, cosiderFigure 1 . . I this gure, the base B C of isosceles ABC has bee exteded to a arbitrarypoit D beyod C The two triagles ADC

A

B D

Figre 1.6

ad AD B are clearly ot coguet because DB DC, ad yet the triagles agreei sidesideagle sice AB AC, AD AD, ad D DThere is oe case where SS is a valid cogruece criterio: whe the agle isa right agle This is the hypoteusearm" criterio, abbreviated ecall tha thelogest side of a right triagle, the side opposite the right agle, i s called the hypotnuof the triagle The other two sides of the triagle are ofte called its arm

(1.3) THEOREM. two right triangles have equal hypotenuses and an arm of oneof the triangles equals an arm of the othe then the triangles are congruent


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10 CHAPTER THE BASCS

B

Figre 1.7

Proof. We are give triagles ABC ad D with right agles at C ad WekowthatAB = DadAC = D,adwewatto showthatABC DMove D, ippig it over if ecessary, so that poits A ad D ad poits Cad coicide ad the diagram resembles Figure 1 7

I Figure 1 7 we have B C = B C A + D = 90 + 90 = 180 adthus B C is a lie segmet, which we ca ow call B Sice A B = D, we seethat AB is isosceles with base B Thus altitude AC is a media by Theorem 1 ad hece BC = The desired cogece ow follows by SSS

We ed this sectio with yet aother proof of the pos asiorum This oe is amusigad very short, but it is somewhat tricky ad should be read carefully

Proof of pon ainorum. We are give isosceles ABC with base B C, ad we watto show that B = C We have AB = AC ad AC = AB Sice also A = A,we ca coclude that ABC ACB by SS It follows that B C sicethese are correspodig agles of the cogruet triagles

xercses B rove the coverse of the pos asiorum Show, i other words, that if B = Ci A BC, the AB = AC

If the altitude from vertex A i ABC is also the bisector of A , show thatAB = AC

If the altitude from vertex A i ABC is also a media, show that AB = AC This fact will be used later, so please do this problem

I Figure 1 8 medias BY ad CZ have bee draw i isosceles ABC withbase BC Show that BY = CZ

sig Figure 1 8 agai, assume ow that BY ad C Z are agle bisectors ofisosceles ABC with base BC Show that BY = CZ

gai i Figure 1 8 assume that ABC is isosceles with base BC, but this time,assume tha BY ad C Z are altitudes Show that BY = C Z ssume for this problem that, as i the gure, the two altitudes actuallylie iside the triagle but observe that this does ot always happe


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A

ANGLES AND PARALLEL LNES 11

C

Figure 1.8

Now i Figure 1 . assume that BY ad CZ are equal altitudes Show thatAB AC

eferrig agai to Figure 1 . let P b e the poit where BY ad C Z meet ssumethat BY CZ ad PY PZ Show that AB AC

Oce more i Figure 1 . let P be the itersectio poit of BY ad C Z s sumethat A B A C ad B Y C Z We ask if it is ecessarily te that P Y P Zou are asked, i other words, either to prove that P Y P Z or else to showhow to draw a couterexample diagram where all of the hypotheses hold butwhere P Y ad P Z clearly have differet legths We drew Figure 1 so that P Yad P Z actually are equal, ad so if a couterexample exists , it would ecessarilyhave to look somewhat dieret from the gure

C Angles and Parallel Lines tranvra is a lie that cuts across two give lies sually, the two give lies areparallel whe we use the word transversal but we do ot absolutely isist o this IFigure 1 .9 for example, lie t is a trasversal to lies a ad b

There is some omeclature that is useful for discus sig the eight agles that we avelabeled with 1 through i Figure 1 . 9 gles that are o the same side of the trasversalad o correspodig sides of the two lies a ad b are said to be corrpondin alesI Figure 19 therefore, L 1 ad L 5 are correspodig agles, as are L ad L ad alsoL3 ad L7 ad, of course, L4 ad L. It is a theorem that correspodig agles are equalwhe a trasversal cuts a pair of parallel lies, ad coversely, if i Figure 19 ay oe

a4 3

b 8 7

Figure 1.9


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of the equalities L 1 L5 L L L3 L7 or L4 L8 is kow to hold, the it is atheorem that lies a ad b are parallel, ad thus the other three equalities also hold

airs of agles such as L4 ad L or L3 ad L5 that lie o oppos ite sides of thetrasversal ad betwee the two give lies are called atrnat intrior agles adpairs such as L 1 ad L 7 or L ad L 8 that lie o opposite sides of the trasversal adoutside of the space betwee the two parallel lies are atat xtrior agles It isa theorem that alteate iterior agles are equal ad that alterate exterior agles areequal whe a trasversal cuts two parallel lies It is also true that, coversely, if iFigure 1 9 ay oe of the equalities L 1 L7 L L8 L3 L5 or L4 L is kowto hold, the lies a ad b must be parallel, ad thus the other three equalities also hold,as do the four equalities metioed i the previous paragraph

We recall also that whe two lies cross, as do lies a ad t i Figure 1 9 theL 1 ad L3 are said to be vrtica an, as are L ad L4 ertical agles , of course,are always equal While reviewig omeclature for agles , we metio that two agleswhose measures sum to 180 are said to be uppmntary, ad if the sum is 90 theages are compmntary Of course, a agle of 180 is a traiht an, ad a agleof 90 is a riht an I Figure 1 9 we see that L 1 ad L4 are supplemetary, ad soif a ad b are parallel, the L 1 L5 ad it follows that L4 ad L5 are supplemetary,as are L3 ad L

We ca apply some of this to the agles of a triagle Give ABC exted side BCto poit D as show i Figure 1 10 I this situatio, LACD is said to be a xtrioragle of the triagle at vertex C, ad the two agles L A ad L B are the rmot iterioragles with respect to this exterior agle

THEOREM. An exterior angle of a triangle equals the sum of the two remoteinterior angles Also the sum of all three interior angles of a triangle is 1 80

Figre 1.10

Proof. We eed to show i Figure 1 1 0 that LACD LA + LB . raw a lie CPthrough C ad parallel to AB as show Now LA LACP sice these are alteate

iterior agles for parallel lies AB ad PC with respect to the trasversal A C lso, L B L PC D sice these are correspodig agles It follows that LACD LACP + LPCD LA + B , as required

We see i Figure 1 10 that LACD+ LAC B LBCD 180 The substitutioof LA + L B for LAC D i this equatio yields the desired coclusio that the sumof the three iterior agles of ABC is 180

Observe that our proof that the iterior agles of a triagle total a straight aglerelies o the fact that it is pos sible to draw a lie through C parallel to A B I fact,


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I ANGLES AND PALLEL LNES 1

there does ot exist a proof of this result that does ot, omehow, deped o paallellies This is because i oEuclidea geometries, it is ot true that the agles of atriagle must total 1 80 , ad yet the oly fudametal differece betwee Euclidea adoEuclidea geometries is i the parallel postulate

triagle, of course, is a polygo with three sides We digress to cosider thequestio of how to d the sum of the iterior agles of a polygo with sides , where may be larger tha thee

Fid a formula for the sum of the iterior agles of a goosider, for example, the case 6 I Figure 1 1 1 , we see two 6gos, which

are usually called hexagos I the left hexago, we have draw the three diagoalsfrom vertex A ( diaona of a polygo is a lie segmet joiig two of itsoadjacet vertices ) I geeral, a go has exactly - 3 diagoals termiatigat each of its vertices , ad this gives a total of ( - 3 /2 diagoals i all (oyou see why we had to divide by 2 here?) polygo is convx if all of its diagoalslie etirely i the iterior The iterior agles of a polygo are the agles as seefrom iside, ad for a covex polygo such as the left hexago i Figure 1 1 1 , theseagles are all less tha 1 80

AD

Figre 1.11

I the right hexago of Figure 1 1 1 , which is ot covex, we see that two ofits six iterior agles exceed 180 ( agle with measure lager tha 180 issai to be a rx agle ) I geeral, a polygo is covex if ad oly if its iterioragles all measure lss tha 1 80 If som itrior agle of a go is exactly equalto 1 80, we do ot cosider it to be a covex go, but i this si tuatio, there aretwo adjacet sides that together form a sigle lie segmet, ad the go ca becosidered to be a ( 1) go iewed as a ( 1 )go, the give polygo maybe covex

Suppos w have a covx go such as hxago ABCDF of Figure 1 1 1Fix some particular vertx A ad draw the 3 diagoals from A This divides theorigal polyg ito exctly 2 triagles ad it should be clear that the sum ofall th trior agls of all of ths triagls is xactly the sum of all iterior aglesof th origial polygo It follows that th sum of the iterior agles of a covexpolygo is xactly 1 80( 2) dgrs

W have ot yt fully solve roblem 1 5 , of course, because we have olycosiderd covex polygos ctually, it is ot quite ecessary for the polygo tob covx to mak th prvious argumt work What is really required is that


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14 CHAPTER THE BAS CS

there should exist at least oe vertex from which all of the diagoals lie iside thepolygo ecall that the deitio of a covex polygo requires that all diagoalsfrom all vertices should be iterior It is coceivable that every polygo has atleast oe vertex from which all diagoals are iterior, but ufortuately, that isot true careful ispectio of the right hexago i Figure 1 1 1 shows that it isa couterexample; at least oe diagoal from every oe of its vertices fails to beiterior It is true, but ot easy to prove, that every polygo has at least oe iteriordiagoal, ad it is possible to use this hard theorem to costruct a proof that forevery go, the sum of the iterior agles is 1 80 ( 2) degrees

There is aother way to thik about roblem 15 that may give additoalisight Imagie walkig clockwise aroud a covex polygo, startig from somepoit other tha a vertex o oe of the sides Each time you reach a vertex, youmust tu right by a certai umber of degrees If the iterior agle at the th vertexis fk the it is easy to see that the right tu at that vertex is a tu through precisely1 80 - fk degrees Whe you retu to your startig poit, you will be facig ithe same directio as whe you started, ad it should be clear that you have turedclockwise through a total of exactly 360 I other words,

L (1 80 - f 360k=

Sice the quatity 1 80 is added times i this sum ad each quatity f is subtractedoce, we see that 1 80 - L fk 360, ad hece L fk 1 80 - 360 180( -2) This provides a secod proof of the formula for the sum of the iterior agles of acovex go

ow importat is covexity for this secod argumet? If whe walkig clockwise aroud the polygo you reach the th vertex ad see a reex iterior agle

there, you actually tu left, ad ot right I this case, your left tu is easily see tobe through exactly f - 180 degrees , where, as before, f is the iterior agle at thevertex If we view a left tu as beig a right tu through some egative umber ofdegrees , we see that at the th vertex we are turig right by 1 80 - fk degrees, adthis i s true regardless of whether fk < 1 80 as i the covex case or f 1 80 at areexagle vertex It is also clearly true at a straightagle vertex, where fk 180The previous calculatio thus works i all cases, ad it shows that 1 80 ( - 2) is thesum of the iterior agles for every polygo, covex or ot

ID Parallelogramsmog polygos, perhaps parallelograms are secod i importace after triaglesecall that, by deitio, a paraoram is a quadrilateral AB CD for which AB CDad A D B C I other words, the oppos ite sides of the quadrilateral are parallel It isalso true that the opposite sides of a parallelogram are equal, but this i s a cosequeceof te assumptio that the opposite sides are parallel; it is ot part of the deitio

(1.6) Opposite sides of a parallelogram are equal


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D PARALLELOGRAMS 15

Proof. I Figure 1 1 2, we are give that A B C D ad AD B C, ad our task is to showthatAB CDadAD BC raw diagoal BD ad ote that A BD CDBsice these are alterate iterior agles for the parallel lies A B ad CD, adsimilarly, D BC A DB Sice BD BD, we see that DAB BCD byS, ad it follows that AB CD ad AD BC, as required

Figre 1.12

D C

Figre 1.13

There are two useful coverses for Theorem 1 6

B

In quadrilateral ABCD suppose that AB CDandAD BCThen ABCD is a parallelogram.

In quadrilateal ABCD suppose that AB CD and AB CD Then ABCD is a parallelogram.

We leave the proofs of these two theorems to the exercises

A quadrilateral is a parallelogram and only its diagonalsbisect each othe

Proof. s show i Figure 1 1 3 , we let X be the poit where diagoals AC ad B Dof quadrilateral A BCD cross Suppose rst that X i s the commo midpoit of liesegmets AC ad BD The AX XC ad BX XD ad also A XB CXDbecause these are vertical agles It follows that AX B CX D by SS, adthus AB CD Similarly, AD BC, ad hece ABCD is a parallelogrm byTheorem 17

oversely ow, we assume that ABCD is a parallelogram, ad we show thatX is the midpoit of each of the diagoals A C ad B D We have B A X XC Dad ABX CDX because i each case, these are pairs of alteate itrior

agles for the parallel lies A B ad CD lso, A B C D by Theorem 1 6 , adthus ABX CDX by S We deduce that AX CX ad BX DX, asrequired

Observe that by Theorem 1 7 , a quadrilateral i which all four sides are equal mustbe a parallelogram Such a gure is called a rhombu I the case of a rhombus thediagoals ot oly bisect each other, but they are also perpedicular lthough this factis ot difcult to prove directly, we prefer to derive it as a cosequece of the followigmore geeral result


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(1.10) iven a line segment BC the locus of all points equidistantfrom B and C is the perpendicular bisector of the segment

Proof. We eed to show that every poit o the perpedicular bisector of BC isequidistat from B ad C ad we must also show that every poit that is equidistatfrom B ad C lies o the perpedicular bisetor of B C

ssume that AX is the perpedicular bisector of BC i Figure 1 14 This measthat X is the midpoit of BC ad AX is perpedicular to BC I other words, weare assumig that AX is s imultaeously a media ad a altitude i ABC adwe wat to deduce that AB AC This is precisely Exercise 1 3 , ad we will otgive the proof here

ssume ow that A is equidistat from B ad C i Figure 1 14 ad draw medaAX of ABC Sice AB AC this triagle is isosceles, ad thus by Theorem 1 2 ,media AX is a lso a altitude I other words, AX is the perpedicular bisectorof B C ad of course, A lies o this lie

A

B CX

Figre 1.14

(1.11) The diagonals of rhombus ABCD are perpendicula

Proof. Sice A B AD we kow by Theorem 1 1 0 that A lies o the perpedicularbisector of diagoal B D ad similarly, C lis o the perpedicular bisectr of B D ut AC is the oly li that cotais the two poits A ad C ad thus AC isthe perpedicular bisector of B D. I particular, diagoal A C is prpeicular todiagoal B D

We close this sectio by metioig aother spcial typ of paralllogram a rct

agle y deitio, a rctan is a quadrilateral all of whos agls r right agles It is easy it see that th opposite sies of a rctagl parallel d so a rctagleis automatically a parallelogram W also kow (by roblm 1 , fr xampl) tha thsum of th four agls of a arbitrary quarilatral is 0 hus if w hav ay quadrilateral with all four agls qual each agl must b 90 ad th gur is a rctaglW rmark also that adjact vrtics of a parllelogram hav supplmt itrioragles It follows easily that if o gle of a paralllogram is a right agl th thparallogram must b a rctagl Fially w mtio that a quadrilatrl that is botha rectagle ad a rhombus is by itio a ar


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ID PARALLELOGRAMS 17

xercses _ rove Theorem 1 7 by showig that quadrilateral ABCD is a parallelogram if

AB CD ad AD BC

rove Theorem 1 8 by showig that quadrilateral ABCD is a parallelogram ifAB CD ad AB CD

rove that the diagoals of a rectagle are equal rove that a parallelogram havig perpedicular diagoals is a rhombus

rove that a parallelogram with equal diagoals is a rectagle

I Figure 1 15 , weare give that AB CDad that A D ad B C are ot parallel Show that D C if ad oly ifAD BC HINT raw a lie through B parallel

t oAD Figure 1.15

NOTE: ecall that a quadrilateral is said to be a trapzoid if it has exactly oepair of parallel sides If the oparallel pair of sides are equal, the trapezoid issaid to be ioc

Show that opposite agles of a parallelogram are equal

I quadrilateral ABC D, suppose that AB C D ad B D Show that AB CDis a parallelogram

I quadrilateral ABCD , suppose that A Cad B D (We are referrig

to the iterior agles, of course ) Show that AB CD is a arallelogramHINT: Show that A ad C are supplemetary

Show that the diagoals of a isosceles trapezoid are equal

I Figure 1 6,we are give that AX B ad AY B are isosceles ad share baseAB Show that poits X, Y, ad Z are colliear (lie o a commo lie) if adoly if A ZB is isosceles with base AB

x

B

Figure 1.16


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I Figure 117, poit is equidistat from the vertices of ABC Thispoit is reected i each side of thetriagle, yeldig poits X, Y, ad ZThis meas that X is perpedicular to B C ad that the perpediculardistace from X to B C is equal tothat from to BC Similar assertios hold for Y ad Z rove thatABC XYZ ad that correspodig sides of these two triaglesare parallel

z

C

x

Figure 1.17

HINT: Show that quadrilaterals ZBA ad CAY are rhombuses educethat B Z ad C Y are equal ad parallelNOTE Give ay triagle, there always exi sts a poit equidistat from the threevertices This poit, called the circumcntr of the triagle, does ot always lie

iside the triagle The assertio of this problem is still valid if lies outside ofA B C or eve if it lies o oe of the sides

Each pair of parallel lies i Figure 118 represetsa railroad track with two parallel rails The distacesbetwee the rail s i each of the tracks are equal rovethat the parallelogram formed where the tracks cros s i sa rhombus NOTE The distace betwee the rails is, of course,measured perpedicularly It should be clear that for

Figure 1.18

each track, the perpedicular distace from a poit o oe rail to the other rail iscostat, idepedet of the poit

ecall that the distace from a poit to a lie is measured perpedicularly GiveAB C, show that the locus of all poits iside the agle ad equidistat fromthe two lies B A ad B C is the bisector of the agle

IE Area

We assume as kow the fact that the area of a rectagle i s the product of its legth adwidth The area of a geometric gure is ofte deoted K , ad so the formula for the areaof a rectagle is K = bh where b is the legth of oe of the sides of the rectagle ad his the legth of the perpedicular sides side of legth b is referred to as the base ofthe rectagle, ad the height h is the legth of the sides perpedicular to the base Ofcourse, the base of a rectagle does ot have to be at the bottom; it ca be ay side

Now suppose that we have a parallelogram that is ot ecessarily a rectaglega we desigate oe side of the parallelogram as the base ad write b to deote itslegth ut this time, the height h is the perpedicular distace betwee the two parallel


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b

II

h iI

b

x b

x

Figure 1.19

E AREA 19

sides of legth b Of course, if the parallelogram happes to be a rectagle , the thisperpedicular distace is the legth of the other two sides

O the left of Figure 119, we see a rectagle with base b ad height h . O theright, we have draw a parallelogram that also has base b ad height h ad we claimthat the parallelogram ad the rectagle have equal areas To see why this is true witha very iformal argumet, drop perpediculars from oe ed of each of the sides oflegth b of the parallelogram to the extesios of the opposite sides , as show i thegure What results is a rectagle with base b + x ad height h where, as show, xrepresets the amout that the base of the parallelogram had to be exteded to meetthe perpedicular The area of this rectagle is (b + x )h ad to obtai the area o theorigial parallelogram, we eed to subtract from this the area of the two right triagles learly, the two right triagles pasted together would form a rectagle with base x adheight h ad so the total area of the two triagles is x h The area of the parallelogramis therefore (b + x)h - xh bh , ad so the formula K bh works to d areas ofarbirary parallelograms ad ot just of rectagles

other way to see why the rectagle ad parallelogram i Figure 1 1 9 have equalarea is from the poit of view of calculus Imagie slicig each of the areas itoiitesimally thi horizotal strips of legth b . We see that each of the two areas isgive by the same formula

K lhbdY ,ad thus the two areas are equal , as claimed I fact, if we carry out the itegratio, weget the formula K bh , as we expect

Next, we cosider areas of triagles I the left diagram of Figure 1 20, w havedraw a triagle with base b ad height h where ay oe of the three sides ca be viewedas the base ad the legth of the altitude draw to that side is the correspodig heghtemember that this altitude may lie outside of the triagle

B&.. CX

Figure 1.20


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To compute the area K of the give triagle, we have costructed a parallelogramby drawig lies parallel to our base ad to oe of the other sides of the triagle, asshow i the gure This parallelogram has base b ad height h , ad so its area is bhThe parallelogram is divided ito two cogruet triagles by a diagoal, ad so the areaof each of these triagles is exactly half the area of the parallelogram (The two triaglesare easily see to be cogruet by SSS ) Sice the origial triagle i s oe of the twohalves of the parallelogram, we see that K bh ad this i s the basic formula for thearea of a triagle

s is idicated with ABC o the right of Figure 1 .20 , ay oe of the three sidescoud have bee desigated as the base, ad for each, there is a correspodig altitude(It i ot a coicidece that the three altitudes are cocurret; we shall see i hapter 2that this is guarateed to happe ) We thus have three dieret formulas for the the areaKAB of ABC We have

AXBCKAB 2

BY AC2

CZAB2

ad we ca deduce useful iformatio relatig the legths of the sides ad altitudes ofa triagle For example, i Exercise 1 .6 , you were asked to show that i a iso scelesABC with base BC, the two altitudes BY ad CZ are equal, ad you were allowedto asume that the altitudes were iside the triagle y the area foula we have justderived, we kow that B Y A C C ZA B sice each of these quatities equals twice thearea of the triagle We ca cacel the equal quatities AC ad AB to obtai B Y C Z,as desired This proof, furthermore, works idepedetly of whether or ot the altitudeslie iside the triagle We remark also that the coverse is true: If we are give thataltitudes B Y ad CZ are equal, it follows from the foula BY AC CZ A B that

AB

AC (This coverse appeared as Exercise 1 .7 .)We ca get some additioal ifoatio about areas of triagles by usig a ittleelemetary trigoometry I Figure 1 .2 1 , we have followed custom ad used the symbolsa , b ad to deote the legths of the sides of ABC opposite vertices A, B, ad C,respectively lso , we have draw the altitude of legth h from A (I the gure, thisaltitude happes to lie outside the triagle, but this is irrelevat to our calculatio) Wehave, of course, K ah , where, as usual, K KAB represets the area of thetriagle We see that si (C) hjb, ad hece h b si(C) , ad if we substitute thisito the area formula K ah, we obtai K ab si(C) Similarly, we also haveK a si(B) ad K b si(A)

It is ow clear that for ay triagle, the equatiosab si(C) b si(A) a si(B)

mus always hold If we divide tough by the quatity abc ad take reciprocals, weobtai the socalled law of sies :

a bsi(C) si(A) si(B)


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A

h

c

B

Figure 121

A

IE AREA 21

B"-

Ca

Figure 122

There is a simple proof of a more powerful formula called the exteded law of sies thatwe prove i hapter 2 other applicatio of the sie formula for the area of a triagleis the followig

THEOREM. et AX be the bisector of LA in 6ABC. ThenBX AB

XC AC

In oher words X divides BC into pieces proportiona to the engths of the nearersides of the triange.

Proof. I the otatio of Figure 1 22, we must show that u/ = /, where ad care as usual ad u ad are the legths of B X ad XC as show To see wh this

equatio holds, let h be the height of 6ABC with respect to the base BC . The his also the height of each of 6A B X ad 6A C X with respect to bases B X ad XCrespectivel We have

uh cx si() = KABX =

2 2ad

h bx si() = KAx = 2 2

where x = AX ad we have writte = LA . ivisio of the rst of theseequatios b the secod ields the desired proportio

s a applicatio , we have the followig, which should be compared with Exercises 1 2 ad 13 like those exercises , it it difcult to see a direct ad elemetarproof via cogruet triagles for this fact

PROBLEM. Suppose that i 6ABC the media from vertex A ad the bisector of LA are the same lie Show that AB = AC.

oution. I the otatio of Figure 1 22, we have u = sice the agle bisector AX isassumed to be a media It follows from Theorem 1 1 that a = , as required


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22 CHAPTER THE BASC S

We have ow give two dieret types of formula for the area of a triagle : oeusig oe side ad a altitude ad the other usig two sides ad a agle Sice the SSScogruece criterio tells u s that a triagle is determed by its three sides , we mightexpect that there should be a ice way to compute the area of a triagle i terms of thelegths of its sides There is I hapter 2, we prove the followig formula, which isattributed to ero of lexadria (c 50 A D :

KABC=l (

a) (s- b) (s- c),

where s (a + b + c) /2 is called the miprimtr of the triagle Of course, a, b ,ad c retai their usual meaigs

We close this sectio with what seems to be a amazig fact

(1.14) s show i Figure 1 23 , poits P Q, ad R lie o the sides ofA B C oit P lies oe third of the way from B to C poit Q lies oe third of theway from C to A ad poit R lies oe third of the way from A to B ie segmetsA P B Q, ad CR subdivide the iterior of the triagle ito three quadrilaterals ad

four triagles , as show, ad we see that exactly oe of the four small triagles haso vertex i commo with ABC rove that the area of this triagle is exactly oeseveth of the area of the orgial triagle

oution. We eed to compute the area Kxyz i Figure 1 23 y choosig uits appropriately, we ca assume that KAB 3 ad we write k to deote KByP We drawlie segmet Y C ad start computig areas

Sice CYP has the same height as 6BYP but its base PC is twice as log, wededuce that Kyp 2k Similarly, sice A Q 2QC , we see that KABQ 2KBQ ,ad thus K B Q KAB 1 This eables us to compute that K YQ 1 KBY 1 3. y the usual reasoig, KAYQ

2KYQ

2 - 6k, ad we kow thatKABQ KAB 2 It follows that KABy 2 - KAYQ 2 - (2 6k) 6kowever, KABP KAB 1, ad thus KByp 1 - 6k ut we ow thatKByP k, ad hece 1 - 6k k ad k 1/7

Simlar reasoig shows that KARX 1/7 KQz ad sice KAR K AB 1 , we deduce that the area of quadrilateral AXZ Q is 1 2/7 5/7Fially, we recall that KAYQ 2-6k 8/7, ad it follows that Kxyz 8/7 5 /7 3/ 7 This is exactly oe seveth of the area of the origial tragle, as desired

A

Ba.CP

Figure 123


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I CRCLES AND ARCS 23

xercses raw two of the medias of a triagle This subdivides the iterior of the triagle

ito four pieces : three triagles ad a quadrilateral Show that two of the threesmall triagles have equal area ad that the area of the third is equal to that ofte quadr lateral

arbitrary poit P is chose o side BC of ABC ad perpediculars PUad P V ae draw from P to the other two sides of the trigle (It may bethat or ies o a extesio of AB or AC ad ot o the actual side of thetriagle This ca happe, for istace, if LA is obtuse ad poit P is very earB or C.) Show that the sum P U + P V of the legths of the two perpedicularsis costat as P moves alog B C. I other words, this quatity is idepedetof the choice of P

Sice a triagle i s determied by aglesideagle, there should be a foula forKA BC expressed terms of a ad LB ad LC . erive such a formula

et P be a arbitrary poit i the iterior of a covex quadrilateral raw the lisegmets joiig P to the midpoits of each of the four sides, thereby subdividigthe iteror ito four quadrilaterals Now choose two of the small quadrilateralsot havig a side i commo ad show that the areas of these two total exactlyhalf the area of the origial gure

IF Circes an Arcs

s all readers of this book surely ow, a circ is the locus of all poits equidistat

from some give poit called the cntr The commo distace r from the ceter to thepoits of the circle is the radiu, ad the word radius is also used to deote ay oe ofthe lie segmets joiig the ceter to a poit of the circle chord is ay lie segmetjoiig two poits of a circle , ad a diamtr is a chord that goes through te ceterThe legth d of ay diameter is give by d 2r ad this is the maximum of the legthsof al chords Fially, we metio that ay two circles with equal radii are cogruet,ad ay poit o oe of two cogruet circles ca be made to correspod to ay poito the other circle

Just as ay two poit deermie a uique le, it is lso true tat ay three poits,uless they happe to lie o a lie, lie o a uique circle

Tee is exacty one circe through any three given noncoinearpoints

Proof. all the poits A B ad C. Sice by hypothesis, there is o lie thoughthese poits we ca be sure that we are dealg with three distict poits , ad wedraw lie segmets A B ad A C. et ad c be the perpedicular bisectors of thesesegets ad observe that lies ad c caot be parallel because lies A B ad A Care either parallel or are they the same lie (Sice A B ad A C have poit A i


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commo, they surely are ot parallel ; they caot be the same lie because we areassumig that o lie cotais all three poits A B, ad C.)

et P be the poit where les ad meet Sice P lies o the perpedicularbisector of AB we kow by heorem 1 10 that P is equidistat from A ad B. Iother words, P A PB Similarly, sice P lies o lie we deduce that P A PCIf we let r deote the commo legth of the three segmets P A PB, a PC wesee that the circle of radius r cetered at P goes through the three give poit

o see that our three poits caot lie o ay other circle, we could appeal tothe obvious" fact that two dieret circles meet i at most two poits, but it isalmost as easy to give a real proof If a circle cetered at some poit Q, say, goesthrough A B , ad C the Q is certaily equidistat from A ad B , ad hece byheorem 1 10, it lies o the perpedicular bisector of segmet A Similarly, wesee that Q lies o lie ad thus Q P because P is the oly poit commo tothe two lies S ice the distace P A r , it follows that the oly circle through AB, ad C is the circle of radius r cetered at P .

Give ABC the uique circle that goes through the three vertices is called thecircumcirc of the triagle , ad the triagle is said to be incribd i the circle Moregeerally, ay polygo all of whose vertices lie o some give circle is referred toas beig incribd i that circle, ad the circle is circumcribd about the polygolthough every triagle is iscribed i some circle, the same caot be said for goswhe 3

he statemet of heorem 1 1 5 would be eater if we did ot have to deal with theexceptioal case where the three give poits are colliear If we were will ig to say thata lie is a certai kid of circle (which, of course, it is ot), we could the say that everychoice of three distict poits determies a circle It is sometimes coveiet to pretedthat a lie is a circle with iite" radius, but of course, this should ot be take tooliterally

We retu ow to our study of geuie circles wo poits A ad B o a circle dividethe circle ito two pieces , each of which is called a arc We write AB to deote oeof these two arcs , usually the smaller s this i ambiguous , a threepoit desigatiofor a arc is ofte preferable I Figure 1 24, for example, the smaller of the two arcsdetermed by poits A ad B would be desigated ad the larger is YB heambiguity i the otatio A B is related to a similar ambiguity i the otatio for aglesFor example, if we write LAO B we geerally mea the agle that i Figure 124icludes poit X i its iterior; we do ot mea the reex agle, with Y its iterior

ow big i s a arc? he most commo way to discus s the size of a arc is i terms ofthe fractio of the circle it is , where the whole circle is take to be 360 or 2 radias arc extedig over a quarter of the circle, therefore, i s referred to as a 90 arc , adwe would write A 90 or A /2 radias i this case Of course, this sizedescriptio for a arc is meaigful oly relative to the circle of which it is a part If weare told, for exampl, that we have two 90 arcs , we caot say that they are cogruetor that they have equal legth uless we kow that these are two arcs of the same crcleor of two circles havig equal radii o remid us that the umber of degrees (or radias)that we assig to a arc gives oly relative iformatio, we use the symbol , which


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B

y

Figure 1.24

CRCLES AND ARCS 25

A

p

Figure 1.25

we ead as equal n degees (o radans) ," and we avod te use of n ts contextI nfoally, oweve, we do speak of equal acs , but t s pobably best to avod tspase except wen we know tat te two acs ae n te same o equal ccles

Gven an ac AB on a ccle centeed at pont , we say tat LAOB s te cntraan corespondng to te ac Snce a full ccle s 360 of ac and one full otaton

s 360 of angle, t sould be clea tat te number of degees n te measure of centalangle LAO B s equal to te numbe of degees n 90 angle at te centalpont fo example, cuts o a quate ccle, wc s a 90 ac Te standad jagonfo te pase cuts o" n te pevous sentence s ubtnd In geneal, we can wteLAOB B: cental angle, teefoe, s equal n degees to te arc t subtends Wecan also say tat te ac s maurd by te cental angle

In a gven ccle , te angle foed by two cods tat sae an endpont s called anincribd angle Some of te nscbed angles n Fgue 1 25 , fo example, ae LAPB,LAQB and LARB

(1.16) An inscribed ange in a circe is equa in degrees to one ha itssubtended arc. Equivaenty the arc subtended by an inscribed ange is measuredby twice the ange.

In Fgue 125, fo example, LA P B and also LAQB Inpartcula, we can deduce tat LA P B LAQB and n geneal, any two nscbedangles tat subtend te same ac n a ccle ae equal s we sall see, ts povdes auseful tecnque fo povng equalty of angles ow consde LARB n Fgue 125ke LAPB and LAQB ts too s an nscbed angle foed by cords troug Aand B , but we cannot conclude tat LARB s equal to te ote two angles because tsubtends te ote ac detemned by ponts A and B In fact, LARB

PB Snceand togete consttute te wole ccle, t follows tat1 1

LARB + LAPB (APB + ARB) 360 1 80 2 2

and tus LARB and LA QB ae supplementay Ts poves te followng coollay toTeoem 1 1 6 ecall tat a polygon s nscbed n a ccle f all of ts vertces le onte ccle


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(1.17) COROLLARY. Opposite anges of an inscribed quadriatera are suppementa

B

A

pp

Q

p

QB

Figure 1.26

Proof of Thorm 1.16. Gven APB nscbed n a ccle centeed at pont tetree cases we need to consde ae llustated n Fgue 1 .26 It may be tat pont falls on one of te sdes of APB, as n te left dagam ltenatvely, mgt len te nteo of te angle, as n te mddle dagam of Fgue 1 .26, o t mgt be

exteo to te angle, as n te gt dagamSuppose st tat les on a sde (say, P B) . aw adus A and obseve tatP s sosceles wt base A P , and so by te pons asnoum, A P entalAB s an exteo angle of A P, and ence t s equal to te sum of te twoemote nteo angles by Teoem 1 .4. Tus

A o A B A + P 2 P ,and ence P as equed

ow assume tat te cente of te ccle les n te nteo of te angle anddaw damete P Q, as n te mddle dagam of Fgue 1 26 y te part of te

teoem tat we ave aleady poved, we know tatAPQ Q and

ddng tese equaltes gves AP B AQB as equedFnally, we can assume te stuaton of te gt dagam n Fgue 1 .26. gan

we daw damete P Q and we get te same two equaltes as n te pevous caseTs tme, subtacton yelds te desed esult.

Imagne te followng expement Mak a lage ccle on te gound and eect two

vertcal poles at ponts A and B on te ccle ow stand somewee else on te ccleand obseve ow fa apart te two poles appea to be, as seen fom you pespectveTe appaent sepaaton of te poles s detemned by te angle fom one pole to youeye and back to te ote pole . Snce ts angle s nscbed n te ccle, Teoem 1 1 6guaantees tat, as we walk along te ccle , te appaent sepaaton of te poles emansuncanged, as long as we stay on one of te two acs detemned by A and B Teangula sepaaton we see fom anywee on te ote ac detemned by A and B,oweve, s te supplement of ts angle


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Suppose now tat te lne toug te two poles uns due not and sout We knowtat fom all ponts on te ccle and east of te poles, te sepaaton between te polesappeas constant. In Fgue 1 .27 , n ote wods, LAP B s ndependent of te coce ofte pont P on te easte ac of te ccle

L

X

v

Figure 127

Wat appens f we eman east of te poles but move outsde of te ccle, topont X, fo example? ommon sense tells us tat snce X s fate fom te poles tans P, te angula sepaaton sould decease We can quantfy ts because, as we see n

te gue, LA PB s an exteo angle of AP X. Tus LAX B LAP B - LXAP , andte appaent decease n angula sepaaton as we move fom P to X s pec sely equalto LXAP . lso, f we notce tat LXAP LAP U we see tat

o 1 LAXB (AB - P) .

lne segment (suc as A X and B X) tat extends a cod beyond a ccle s calleda cant, and so we ave tus poved te followng esult.

(1.18) COROLLARY. The angle between two secants drawn to a circle from an

exterior point is equal in degrees to ha the derence of the two subtended arcs.

etunng now to ou two poles, we see n Fgue 1 .27 tat f we move fom pot Pto a pont nsde te ccle and east of te poles, te appaent sepaaton of te polesnceases fom L AP B to L A B. Snce LA B s an exteo angle of A P, t followstat te amount of ncease equals LA B - LAPB L P A Tus

We ave poved te followng

o 1 LAB (AB + PV ) .2

(1.1) COROLLARY. The angle between two chords that intersect in the inte rior ofa circle is equal in degrees to ha the sum of the two subtended arcs.

We ave seen tat te pat of te plane east of ou two poles s subdvded nto teesets . Eveywee on te ccle, te angula sepaaton of te poles s Insde teccle, te angula sepaaton s always geate tan ts quantty, and outsde te ccle,te angula sepaaton s always smalle We close ts dscus son wt a lttle execse


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28 CHAER THE BAS CS

Given two points and some angle sketc te locus of all points on te plane fromwic te angular separation of te two given points is equal to ons ider separatelyte tree cases < 90, 90, and 90.

(1.0) 6foottall rectangular painting is ung ig on a wall, wi itsbottom edge 7 feet above te oor n lover wose eyes are 5 feet above teoor wants as good a view as possible, and so se wants to maximize te angularseparation from er eye to te top and te bottom of te painting ow far from tewall sould se stand?

T

M b

ep

Figure 1.28

oution. In Figure 128, orizontal line e represents te possible positions of teviewer's eye, 5 feet above te oor ine B is te wall on wic te picture isung, and and B represent, respectively, te top and bottom of te picture. We

seek a point P on line e tat maximizes L B P ltoug tis can be done bycalculus tecniques, we present an easier metodWe kow tat point B is 2 feet above line e and tat is 6 feet iger, or

8 feet above e . Te mdpointM of B is tus at te average eigt (2+8) /2 5 feetabove e raw te perpendicular bisector b of B so tat b is orizontal and 5 feetabove e and coose point on b so tat B 5 raw te circle of radius 5centered at and note tat tis circle is tangent to e at some point P Te circle, inoter words, touces line e at P , but it does not extend below e .

We argue tat tis point P solves our problem Every oter point on line e liesoutside te circle and tus sees" te picture B wit a smaller angle tan does P

In oter words, L B P i s te maximum we seek To answer te question tat wasasked, we need to kow ow far point P is from te wall Tis distance is equalto M, and so we examine te rigt triangle MB We kow tat ypotenuse B 5 because B is a radius of te circle lso, MB 3, and ence by teytagorean teorem, we see tat M 4 Te answer to te problem, terefore,is tat te art lover sould stand 4 feet from te wall

Since we used te ytagorean teorem in te solution to roblem 1 20, perapstis s a good place to digress to give an elegant noncomputational proof


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I CRCLES AND ARCS 9

(121 (Pythaora) Ifa right triangle has arms of lengths a and b andits hypotenuse has length c then a + b c

b

t b b

Figure 1.29

b

b

b

b

Proof. s we sall explain, te wole proof is visible in Figure 1 .29 On te left, wesee our given triangle . Te middle diagram sows a square of side a +b decomposedinto two squares and four rigt triangles . We observe, furtermore, tat eac of terigt triangles as arms of lengt a and b and so eac is congruent to te originaltriangle by SS . Te two smaller squares in te middle diagram ave side lengtsa and b and so te area remaining in te big square of side a + b wen four copiesof our given triangle are removed is exactly a + b

On te rigt, we see anoter square of side a + b decomposed tis time intoone square and four rigt triangles . gain, eac of te rigt triangles is congruentto te given one by SS , and so te side lengt of te smaller square is exactly c It

follows tat te area remaining wen four copies of our triangle are removed froma square of side a + b is c We just saw, owever, tat tis area is equal to a + b and tus a + b c as required.

Tere is one crucial detail we ave omitted. It is clear tat te square" ofside c in te gure on te rigt is a rombus, but wy are its angles equal to 90?ook, for example, at te top of te diagram on te rigt. We see tat te anle ofte rombus togeter wit L 1 and L2 make a straigt angle of 180 In te originaltriangle, owever, we know tat L 1 , L2, and a rigt angle sum to 180, and it followstat te angle of te rombus is 90, as required

We mention an alternative argument tat could be used to see wy te rombus witside lengt c in te rigt diagram of Fiure 1 29 as to be a square. Observe tat tere isa rotational symmetry in te gure In oter words, if we rotate te entire large squareby a quarter turn, wat results is identical to te gure wit wic we started. It followsfrom tis tat all four angles of te rombus must be equal Since we kow tat te sumof te angles of any quadrilateral is 30, we deduce tat eac corner of te rombusis 90, and tus te rombus is a square.

We return now to our study of angles in circles . Tere is a special case of Teorem 1 1 tat is used so often tat it deserves special mention


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(1.2) COROLLARY. Given ABC the angle at ve rtex C is a right angle andonly side A B is a diameter of the crcumcircle

Proof. We ow tat te cicumcicle exists by Teoem 1 5 In tis cicle, AB

i s measued by2LC. Hee, of couse, we have written B to denote the ac not

containingC that these points determine. hord A B is a diameterprecisely when

B

1 80, or equivalently, when LC=

90 . This completes the proof.

eaps we sould comment on te pase if and only if," wic appeas in testatement of oollay 1 22 Te if pat of te statement assets tat A B is a diamete,then LC is a igt angle In ote wods, tis pat of te coollay says tat wenevewe know tat AB is a diamete, we can conclude tat L C is a igt angle Te only ifpat of te coollary tells us tat te only way tat LC can be a igt angle is fo A B tobe a diamete In ote wods, LC is a igt angle, then AB must be a diamete eonly if pat of te assetion, teefoe, is pecisely te convese of te if pat Te if andonly if fom of matematical statements is so common tat tese fou wods ae often

combined into te single abbeviation i," altoug we do not use te abbeviation intis bookGeneally, wen we ae asked to pove an assetion tat says someting in te fo

abc if and only if xz," we ae expected to povide two poofs We pove te if patby assuming xz and someow deducing abc We pove te only if pat by stating allove and assuming abc and deducing Of couse, it doesn t eally te wetewe do if o only if st, as long as bot get done In some exceptional cases , oweve,it is possible to pove bot te if and te only if pats of an if and only if assetionsimultaneously Te poof we just gave fo oollay 1 22 is an example of t

ee is an amusing pactical" application fo oollay 1 22 Suppose a cicle is

printed on a piece of pape, and we want to nd its exact cente If te cicle ad beendawn wit a compass, we could old te pape up to te ligt to nd te tiny oletat would mak te cente, but fo a pinted cicle, we need a dieent metod Takeanote piece of pape, wit a 90 coe, and place it down ove te printed cicle sotat it coves te appoximate location of te cente and so tat its coe lies exactlyon te cicle ow mak te point on te cicle wee eac of te two sides of te igtangled coe cosses te cicle emove te coveing pape and use a staigtedge tojoin tese two maks y oollay 1 22, te line segment tat esults is a diamete ofte cicle epeat tis pocess to daw a second diamete and mak te point wee tetwo diametes coss Tis is clealy te cente of te cicle

Obseve tat given any line segment A B, tee is a unique cicle aving A B asa diamete Tis, of couse, is te cicle centeed at te midpoint of A B and avingadius A B . note way to state oollay 1 22, teefoe, is tat LC is a igt anglein A BC if and only if point C lies on te unique cicle aving side AB as a diameteote tat LC < 90 if C lies outside of tis cicle and LC 90 if C is in te inteioecall tat angles smalle tan 90 ae said to be acut, and angles between 90 and180 ae obtu

ecall tat a line is tannt to a cicle if it meets te cicle in exactly one pointTroug evey point P on a cicle, tee is a unique tangent line, wic is necessarily


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perpendicular to the radius terminating at P One wa to see wh the tangent must beperpendicular to this radius is from the point of view of calculus

In Figure 130we have drawn tangent line PT and we want to compute LPT,where P is the diameter that extends radius P. hoose a point Q on the circle nearP and draw the secant line P Q If we move Q closer and closer to P , we see that LA P Tis the limit of LP Q ut as Q approaches P, we observe tha A Q approach s

180 since Q P is a semicircle It follows that LP Q approaches 90 and so P isperpendicular to the tangent as claimednother fact that we can see in Figure 1 30 is that L QP T between chord P Q and

tangent P T is the complement of LP Q Thus

1 1 1 L QP T = 90 - LAP Q = 90- - A Q = - ( 1 80 - AQ) = - P Q .2 2 2

In oher words we have the following

A s

o

Figure 130 Figure 1.31

(1.) THEOREM. The age betwee a chord ad the taget at oe of its edpoits

is equa i degrees to ha the subteded arc

We leave as an exercise the following consequence

(1.4) COROLLARY. The age betwee a secat ad a taget meetig at a poitoutside a circe is equa i degrees to ha the derece of the subteded arc

Two circles are said to be mutuay tannt at a point P if P lies on both circlesand the same line through P is tangent to both circles This can happen exteall whente two circles are on oppos ite sides of te tangent line or inteally when the circles

are on the same side of the tangent and one circle is inside the other(1.5) PROBLEM. Given two exteall mutuall tangent circles with common

point P, draw two common secants D and BC through P , as in Figure 131Show that B and CD are parallel

oution. Since B C is a transversal to the two lines B and CD it sufces to showthat the alteate interior angles L B and LC are equal raw the common tangen ST

and note that LDPT P D b Theorem 1 23 lso of course LC P D, and


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so it follows that LC LDP Similarly L B LAP ut LDPT and LAP are vertical angles s o they are equal and it follows that L C L B as desired

Exercises IF

F. Suppose that A B and CD are chords on two circles with equal radii Show that

f AB CD if and on AB CDNOTE: The assertion of this problem is really pretty obvious but nevertheless you should provide a proof here

F.2 et A B C and D be placed consecutively on a circle et , X, , and Z

be the mdponts of AB B C CD and DA respectvely Show that chords and X Z are peendicular

F. et P be a point exterior to a circle centered at point and draw the two tangentsto the circle from P et and T be the two points of tangency Show that Pbisects L P and P P

F.4 In the situation of the previous exercise show that 1 80 L P F.S In A B C prove that L A is a right angle if and only if the length of the median

from A to B C is exactly half the length of side B C

F.6 In quadrilateral ABC D assume that LA 90 L C raw diagonals A Cand BD and show that LDAC LDBC

F.7 In the situation of the previous exercise assume that diagonal A C bisects diagonalB D rove that he quadrilateral is a rectangle

F.8 In Figure 1 .32 , two circles meet at points P and Q and diameters P A and P Bare drawn Show that line AB goes through point Q

A

p

Figure 132 Figure 133

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