Geomet for Colege Sdents
I. Martin Isaacs
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saacs, Martin, (date) Geomety for college students Martin saacs p cm ncludes index SBN 0-534-35179-4 (text) 1. Geometry Title 00-056461 QA445 63 2000 516--dc21
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Preface
his book is primariy intended for coege mathematics students who enjoyed high schoo geometry and who wish to ea more about the amaing properties of ines circles triange triangess and other geometric gures I n pparticuar articuar I hope that those who are are preparing preparing to become highschoo high schoo mathematics mathematics teachers teachers wi nd nd inspiration in spiration here here to hep them share their enthusiasm enthusia sm and enjoyment of geometry geometry with their own future future students utt why u why shoud sho ud anyone study geometry? geo metry? One reas reason on of course cou rse is that geometry geo metry and its descendant desce ndant trigo trigonometr nometry y are essentia ess entia toos in engineering architecture navigation and other other discipines hese practica appications however however surey sure y do not expain why it is that for for centurie centuriess geometry has been taught to amost every every student and why at east recenty a person who knew no geometry geometry was not considered to be propery edu until recenty cated I think there there are at east ea st two reaso reasons ns more important important than usefunes us efunesss that expain why geometry has been and shoud continue to be a part of the schoo schoo curricuum Since Eucid some 300 years ago geometry geometry has been taught as a deductive deductiv e science with theorems and proofs s a consequence, generations of geometry students have eaed how to draw vaid concusions from hypotheses and how to detect and avoid invaid invaid reasoning reasoning In other words by studying geometry geometry students students can c an e how to think are other subjects subjects that coud aso be used to teach deductive reasoning reasoning Of course there are but geometry is especiay eective because it seems to have the perfect baance of depth and concretenes s Many of the theorems that we prove prove in geometry are dee in the sense that that they they assert as sert something nonobvious nonobvious and sometimes even surprising hey are are concrete because students can easiy e asiy draw draw the appropr appropriat iatee trianges circes circes or whatever, whatever, and they they can see that what is aeged to happen actuay does appear appear to happen happe n Geometry is aso beautifu and some of its theorems are so amaing as to seem amost mracuou mracuouss In fact fact much the same s ame comment comment coud be made about most areas of 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 dea of mathematica sophi s ophisti stication cation to marve marve at the fact fact that if a ine is drawn through each vertex vertex of any triange triange and if each of these these ines is i s perpendicuar to the opposite oppos ite side si de of the triange triange then these three ines a go through a common point One coud cou d argue that that the fundamenta fundamenta theorem theorem of cacuus cacuu s for exampe is equay equay amaing and beautifu beautifu but unfort unfortunatey unatey it is not possibe poss ibe to appreciate appreciate it without without rst studying cacuus cacu us ut ut the aesthetic vaue of geometry geo metry extends beyond beyon d the strikng statements of its theorems
PREACE
Many Many of the the proofs have their own subte and eegant beauty which whic h unfortunatey is a itte harder harder to perceive. Nevetheess Nevetheess we expect that that most readers of this this book wi ea to enoy the beauty of o f the the proofs proofs as a s we as that of the theorems theorems.. In shot we shoud continue to study and teach geometry because it is a highy attracti attractive ve subect and because b ecause we can c an ea from from it something about deductive reasoning and the nature of mathematica mathem atica proof. It seems se ems cear ce ar therefoe therefoe that as in the past students shoud shou d continue to see theorems theorems being proved in their their geometry geometry cass cas s they should sh ould be taught how to understand proofs and perhaps even more impotant they shoud shou d lea how to invent and write proofs. I have seected se ected for this book b ook some so me of the the more spectacuar theorems in pane ge ometry ometry and I have presented ustications usti cations of thes thesee facts facts using u sing a vaiety of diere dierent nt techniques of proof. Mosty ignored however are the kind of unsurprising theorem which whie required by mode standards of mathematica rigor can seem rather pointess to students students It requires considerabe sophistication to appreciate why why anyone woud want to prove a fact fact that that seems competey competey obvious obviou s . Even professiona professi ona mathe matcians, most of whom do understand the the signicance signi cance of these resuts often nd their their forma forma axiomatic proofs proo fs somewhat du earning proofs proofs shoud not be an unewarding chore instead we expect that students wi demand proofs proofs because becaus e the assertions being estabished are otherwis otherwisee so incredibe. is book was written as a text for oege Geometry which is a course that I have taught severa times at the niversit of Wisconsin Madison. The courses principa audience consi c onsists sts of o f sophomore and unior unior undergradua undergraduate te math maors who are speciaiing in secondary s econdary education. For them the course is required but there there is aso a so a substantia sub stantia minority who take the course eectivey eectivey Some S ome of these students s tudents simpy want to earn geometry whie others take oege Geometry because they nd it to be a more gente gente and accessibe acces sibe introduction to to mathematica proof proof than than a course cours e in abstract abstract agebra or advanced cacuus. cacuus . It is my beief bei ef that for some students s tudents the study of geometry geometry is i s an exceent preparation for these these more difcut difcut abstract courses . I have been dissatised with the avaiabe texts that might be used for suh a course. course . Many Many incude in cude a umber umber of interesting topics topics within a arge arge samping s amping of assorted geometric materia. ut they do not put the focus where I think it beongs: on the reay pretty theorems and their proofs proofs that in my opinion opini on shoud shou d be at the heat heat of a geometry course for coege students. so they often devote much more space than I think think is appopriate appopriate to fomaism fomaism and axiomatics Most Mo st students probaby do not nd this especiay esp eciay exciting excitin g and I share that opinion. here aso exist exi st some s ome wonderfu wonderfu books that that are ed with spectacuar s pectacuar theorems theorems and eegant proofs but none of these seems quite suitabe as a text for this geometry course either I have found that that most mos t students who register for for oege Geometry Ge ometry claim to remember remember very itte from from their one previous previous exposure to geometry in high schoo. schoo . For the sake of this maority some review and accimatiation are necessary. text is needed therefore that starts from a point cose to the beginning and introduces the notion of proof genty and then then gets to the the good stu" stu " as quicky as pos sibe. sibe . ecause ecause I coud not no t nd a book b ook that covered the right materia materia at the right pace pac e and an d that stated stated from the right pace pac e I taught the course cou rse many times without a text.
PREACE
Most of my students seemed to enjoy the course and many of them became very excited about about geometry. In this this book which is an expansion expansion of my course lecture notes I have have tried tried to reproduce as closely clos ely as pos sible the experience experience of the classro cla ssroom om and so so I hope that my readers readers will wi ll also als o nd that geometr geometryy is an enjoyable enjoyable and an d exciting sub s ubject ject Most of all I hope that those of my students and readers readers who wh o are or will be teaching highschoo high schooll mathematics will wil l convey some of that excitement to their their own students students I am grateful to the follo following wing reviewers for their their helpful comments comme nts:: 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
Contents
CHPER 1
1 1 E 1F G 1
Introduction and poogy 1 ongruent rianges rianges 5 nges and and arae arae ines 1 1 araeograms 14 rea 18 irces and rcs rcs 23 oygons in irces irces 34 Simiarit Simi arityy 39
CHPER 2
2 2 2 2 2E 2F 2G 2
5
he ircumcirce 50 he entroid entroid 5 he h e Euer ine, Orthocenter Orthocenter and and Nineoint irce irce 0 omputations 7 he Incirce Incirce 7 3 Exscribed irces irces 8 0 Moreys Moreys heore heorem m 82 Optimization Optimization in rianges rianges 85
CHPER 3
3 3 3 3
4
Simson ines 94 he uttery uttery heorem 105 ross ross atios 10 he adica xis 119
CONTENTS
CHPER 4
v' v 4 4 4 4
eva evass heorem heorem 1 25 Interior Interior and and Exterio Exteriorr evians evians 1 3 1 eva evass heorem heorem and nges 1 3 Meneaus Meneaus heore heorem m 14
CHPER 5
56
V M f f 5 5 5 5 5E 5F
ectors 15 ectors and Geometry 158 ot roducts roducts 1 3 heckerboards 1 it it of o f rigonometr rigonometryy 170 inear Operators 172
CHPER 6
E F
ues of the Game 1 82 econstructing econs tructing rianges rianges 1 87 angents 191 hree ard ard robems 1 9 onstructibe ons tructibe Numbers 203 hanging the ues 208
6
5
GEOMETRY or f
College Students College
CHPER ONE
Te Basics
A
ntroduction ntroduction and Apology
Elemen ts, it has been Since the appearance about about 2300 years years ago of Eucids book b ook The Elements, tradtiona to study geometry as a deductive discipine, using the socaed axiomatic method. Ideay this means that the geometer (or geometry student) starts with a few assumed as sumed facts facts caed ca ed axioms or postuates postu ates and then systematicay system aticay and and carefu carefuyy derives the entire entire content of the subject subject using nothing but pure ogic. ogic . he he axiomatic method requires requires that this content, which for geometry is a coection c oection of o f facts facts about trianges circes circes and other gures shoud be presented as a sequence sequen ce of ever ever deeper deeper theorems theorems each rigorousy rigorou sy proved using usi ng the axioms axio ms and earier earier theorems . Geomet Geometry ry has continued to deveop in the centuries centuries since si nce Eucid. Eucid . Mathematicans Mathematicans incuding many taented amateurs amateurs have discovered a weath of beautifu, beautifu, surprising and even spectacuar properties properties of geometric gures gures and these these together together with the geomet geo metry ry Element s, constitute what is today of The Elements, today caed cae d Eucidean geometry. geometry. mazing as many of these asserted facts may be we can be condent that they are correct because they are proved they are not merey tested by experiment and conrmed by measurement Furthermore ore we do not have to rey on the authority authority of o f Eucid or his and observation. Furtherm success succe ssors, ors, because becaus e with a itte practice and prepar preparation, ation, we can read and undertand undertand these proofs ourseves. With uck we might nd simper and more eegant proofs for some som e of the the known theorems of o f Eucidean Euc idean geometry geo metry.. It is even even possib pos sibe e that we might discover disc over new geometric facts that have never been seen s een before and invent our own proofs for them. geometry has resuted not ony in the discover disco veryy of o f amazing he deveopment of geometry facts, facts, but aso in the invention of new and powerfu powerfu techniques of proof proof and these too are considered consi dered part of Eucidean geometry. geometry. escartes escartes invention invention of coordinate geometry is an exampe of this. third thread thread in the history of geometry, geometry, especiay e speciay in mode times has been an investigation into the foundations foundations of the subject subject.. It has been asked for instance whether or not a of Eucids Eucid s axioms and postuates are reay vaid and whether or not we reay need to assume them a. ut we say amost nothing about these foundationa foundationa issues is sues in this this book. bo ok. Instead, I nstead, we concentrate concentrate on the two more cassica cass ica themes of geometry: facts and proofs. We oer the reader a seection of some of the
2
CHAPTER
THE BASCS
more striking facts of Eucidean geometr ge ometry y and we present enough proof proo f machinery machinery to estabish estabish them. In the centuries since Eucid the accepted standards standards of mathematica mathematica rigor have have su fciency ci ency of proofs proofs in the stye of Eucid has been chalenged. One changed changed and the suf objection for example is that Eucid reied too heaviy on diagrams and that he and the other cassica geometers did not aways prove facts they considered to be obvious from the gures. In this book we foow the tradition of Eucid and of most of his succes sors and so s o we are wiing to t o use some of the information information contained in caref carefuy uy drawn drawn gures g ures.. ut ut we must mus t not rey on diagrams for for certain certain other types of infoation, infoation, and unfort unfortunatey unatey it is i s difcult difcult to be precise about exacty what we are wiling to read read off off from a diagram and what requires a proof. We are are condent cond ent that readers wil catch on to this fairly fairly quicky quick y however how ever and we apoogie for the ambiguity. erhaps an example wi hep carify carify the situation. si tuation. In Figure Figure 1 . 1 ange ange bisectors A B , and C Z have been drawn in ABC hree facts that seem cear c ear in the gure are are : A
Figure 11
1 B < C In other words words line segment B is shorter shorter than ine segment C 2 . he point , where where the bisector of L A meets ine BC, ies on o n the ine segment BC 3 he three ange bisectors are concurrent. efore we discuss these observations et us be clear about the meanings of the technica 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 be parae parae aways meet 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 that the ines are concurrnt he remaining undened technica word in the precedng ist of facts is segment line mnt is that part of a ine that ies between two given points on the ine. Observe that that in assertion assertion ( 1 ) we are are using the notation notation B in two dierent dierent ways: ways : In the inequality, B represents represents the length of the ine segment, se gment, which is a number and then ater B is used us ed to name the segment itsef. In addition the notation B is often used us ed to represent represent the entire ine containing the points B and , and not just ju st the segment s egment they determine. determine. ut we sha always provide enough infoation infoation so that it is cear c ear from from context which of the three three possible pos sible meanings meaning s is intended. For exampe in inequaities or equations it is obvious obvio usy y the numerica numeric a interpretation interpretation that is wanted. Each of ( 1 ) (2) ( 2) and (3) is true but we need to distinguish distingui sh among three di diere erent nt types of truth here. First note that the fact in (1) that B < C is an accident. his
A NTRODUCTON AND APOLOGY
inequality happens to be true in this gure, gure, but it is not an instance of some s ome genera generall or universal truth Even Eve n in a particular particular case, this sort of information can be be unreliable unrelia ble since sinc e it depends on the accuracy of the the diagram and the care with which measurements are made It is never considered con sidered valid to read read o an equality equalit y from from a diagram and it is rare that that one is justi ju stied ed in reading of o ff an inequality Fact (2) that point lies between between points B and on line B , is not not accidenta accidentall Indeed, it seems completely obvious that the bisector of each angle of an arbirary triangle must always intersect the opposite side of the triangle In other words, the bisector intersects the line segment determined by the other two vertices of the triangle triangle Note that that although althoug h this is true about angle bisectors bi sectors,, it can fail fail for other important lines associated asso ciated with a triangle triangle For example example the attud drawn from from A, which is the line through A perpendicular to line B , may not meet the segment B lthough this failure ailure (accidentally) (accidentally) does not happen happen for for AB A B of Figure 1 1 , it certainly certainly can happe happenn for other triangles triangles he obvious" obvious " fact that angle bisectors of triangles triangles meet the opposite sides is the sort of information that Euclid was, and we are willing to read off from diagrams lthough lthoug h it is a fact fact that for every triangle triangle,, the three angle bisec bi sectors tors are concurrent, co ncurrent, as in assertion ass ertion (3) we w e follow follow Euclid in that that we do not consider cons ider this this to be obvious he experimental evidence evid ence of o f even an accurate computerdrawn diagram (or of many many such diagrams) is not sufcient for us to accept this as a universal universal truth we require a proof proof eaders will probably have seen se en a proof of this theorem in highscho highs chool ol geometry, ge ometry, and it is also als o proved here here in hapter hapter 2 hus there is an a n inherent ambiguity ambiguity about which informa information tion can ca n be reliably estab lished from from diagrams and which cannot ecause ec ause of this ambiguity, modern standards of mathematical mathematical rigor require require that there there must be no reliance whatsoever on gures gure s Geome try without without diagrams diagrams was made possible by avid avid ilbert ilbert (1 821 82 1 943) who, building on the work work of his predecessors predeces sors constructed co nstructed an an appropriate appropriate set of precisely precise ly stated axioms from which he was able to prove everythi everything ng formall formallyy and without diagrams Euclid Eu clid s ideal that geometry geometry should shoul d be a purely deductive enter enterprise prise was thus nally nally realized by ilbert at the turn of o f the 20th century In particular ilberts ilbert s axioms allowed him to prove our observation (2), that an angle bisector bis ector of a triangle always meets mee ts the opposite oppos ite side sid e of the triangl trianglee ut ut it is far far from a triviality to prove rigorously, rigorously , and without relying on a diagram, that the bisector bise ctor of B A meets liline B at a point that lies between between B and Further Furthermore, more, there there would be no hope hop e of proving such a thing thing without having a precise denition of the word between, which whic h ilbert was was able to provide provid e researchers into the foundations foundations of geometry geometry he achievements of ilbert and other researchers were substantial and signic sig nicant, ant, but it is my opinion that, that, by far, far, the most interesting theorems of geometry are those that provide surprises We feel, for example, that the fact fact that the three angle bisectors of a triangle are always concurrent is much more exciing than is the fact fact that the the angle bisectors bis ectors meet the opposite sides s ides of the the triangle For his reason we have chosen chose n to present in this book as many unexpected facts facts and surprisin surprisingg theorems theorems as space allows allows Since Sinc e we want to get to these these quickly quickly we must omit ilbert ilbertss careful and rigorous treatment of the foundations foundations of geometry, geometry, and instead, we will follow tradition and rely on diagrams without specif spe cifying ying exactly exactl y the extent of our
4
CHAPTER
THE BASCS
reiance. so since the reader reader of this book wi have studied some geometry g eometry in high schoo schoo we wi not start our presentation presentation at the very very beginning of the the subject. We have aready apoogize apo ogizedd for the ambiguity abou a boutt how much mu ch information we are aowed to obtain obtain from from diagrams. Some apoogy is aso as o appropriate appropriate concerning the the issue is sue of how much highschoo geometry we are assuming. Students may fairy compain when they are doing homework exercises exercises that it is uncear uncea r which facts facts they must prove and which they can c an merey quote as remembered from from schoo s choo . We do not attempt to give a compete ist of assumed resuts resu ts but we sha s ha show by exampe the eve of proof that we expect and we sha devote much of the rest of this chapter to a review of some of the essentia facts denitions and theorems from from highschoo highschoo geometry. efore proceeding with our review of highschoo geometry we discuss briey some of the issues in the foundations of geometry to which we referred earier We mentioned that Eucid caed his unproved assumptions axioms and postuates. he distinction, which is not considered signicant today today is that Eucids Eucid s axioms ax ioms conceed conc eed genera ogica reasoning, whie his postuates were more specicay geometric. For exampe one of Eucid s axioms is things equa equa to the same thing are are equa to each other" other" whereas his parae postuate ess entiay asserts that given a ine and a point not on that ine there exists exis ts one and ony one ine through the given point parae to the given ine." ctuay ctuay Eucid s parae postuate postuate is not stated in precisey precisey this way it appears appears in a somewhat so mewhat more compicated but ogicay equivaent formuation. formuation. Over the centuries, Euclids Euclid s parae postuate has engendered a great dea of interest and controversy. controversy. S omehow the existence and uniqueness of a ine ine parae to a given ine through through a given point seemed see med ess obvious than the facts facts asserted as serted by the the other postuates. Geometers fet uncomfortabe unc omfortabe assung assu ng the parae postuate pos tuate and they attempted instead to prove it they tried to deduce it from from the rest of Eucids Eucid s axioms and postuates efore efore we discuss discu ss these attempts we w e shoud shou d stress that the the parae parae postuate makes two separ s eparate ate assert ass ertions ions each of which woud have to be proved. proved. It woud be neces nec essary sary to show that a parae ine (through the given point) exists, and one woud aso as o need to prove that it is unique ow might a proof of the parae postuate proceed? proceed? One coud ass assume ume that it is fase fase and a nd then try try to derive derive some s ome contradictory contradictory concusions concusion s ssume s sume for exampe exampe that there there is some s ome ine A B and some point P not on AB, and there there is no ine parae to A B through P If by by means of this assumption a ssumption one coud deduce the existence o f a triange triange �Z for which Z and aso < Z, then this contradiction woud prove at east the existence part of the parae parae postuate. What actuay happened when this was tried was that apparent contradictions contradictions were derived derived and the existence of gures that seem impossibe impos sibe was proved. For exampe the assumption that no ine parae parae to A B goes goe s through through point P yieds a triange triange �Z that has three right anges. anges . Surey Surey this is impossibe" imposs ibe" we might migh t say but how can we prove prove this imposs impo ssibiity ibiity?? We We know from from high schoo that the three three anges of a triange triange sum to 1 80 and an easy experiment conrms conrms this fact (ear o the three coers of a paper triange and ine them up to see that the three three anges tota tota a straight ange.) ange. ) It may seem that by proving the existence of a triange with with three rght anges, we have have the desired contradiction contradiction but this is wrong. he highschoo highs choo proof that that L + L + L Z 1 80 utimatey reies reies on the parae parae postuate which we are temporariy temporariy refusin refusingg to assume. as sume. so, so , the experiment with paper trianges trianges is certainy not a mathematica proof.
B CONGRUENT TRANGLES
fter fter repeated attempts to obtain obtain contradiction con tradictionss from from the denia den ia of o f either either the ex istence or the uniqueness part of the parae postuate it was reaized by J oyai N. obachevski and Gauss in the 19th century that athough such denias yied seemingy ridicuous ridicu ous situations situations such suc h as trianges trianges with three right anges anges no proof of a contradiction contradiction was possibe pos sibe.. In fact fact it was proved that no such such proof is possib pos sibe.e. In other words one coud c oud buid a perfecty perfecty consistent consi stent deductive deductive geometry geometry by repacing Eucid s parae parae postuate po stuate with wi th either one one of o f two ateative ateative new new postuates. po stuates. One of o f these denies o f a parae to some s ome ine through through some point poi nt and the other asserts the the existence of existence of at east two such suc h paraes. paraes . Each of the two two types of geometry that arise in this way is said to be nonEucidean and each has its own set s et of proved theorems. theorems. he geometry where where no parale exists exis ts is i s caed eiptic geometry, geometry, and when more than one parae to a ine goes goe s through a point we have hyperboic hyperboic geometry. he deductions of each of the two types of nonEucidean geometry contradict contradict each other and they aso contradict contradict the theorems of cassica cass ica Eucidean geometry geometry but each of the three three avo avors rs of geometry geometry appears to be inteay consistent consi stent and from from a mathematician mathematician s point p oint of view view they are equay vaid. ctuay it is known that if Eucidean geometry is internay consistent then the two nonEucidean geometries are aso consistent but no forma proof proof of the the consi co nsistency stency of Eucidean geometry has been found. he questio que stionn of which if any any of the three three avors of geometry that we have been discussing describes the rea word is interesting, but it is not reay a mathematica question question it beongs beon gs in i n the ream ream of physics. physics . he itte experiment experiment with the paper triange certainy suggests that we ive in a Eucidean universe but this has not been rmy estabished on a arge arge scae. sc ae. Indeed modern theories of the structure structure of the cosmos cos mos incuding Einstein Eins teinss theory of genera reativit reativity y suggest sugge st that none of the three three geometries provides an entirey accurate description of the universe in which we actuay ive. Nevert Neverthees heesss Eucidean Euci dean geometry provides a very good approximation to reaity on a human human scae sc ae and so it is usefu u sefu for for practica purposes such s uch as engineering e ngineering navigation, and architecture.
IB
Congruent Con gruent iangles
We devote the rest of this chapter to some s ome of the basic facts facts and a nd techniques of Eucidean geometry geometry much of which wi be a review for for most readers. eca that two geometri g eometricc gures are conrunt if informay speaking they have the same size and shape. S omewhat more preci precisey sey two gures gures are congruent congruent if one can be subected to a rigid rigid motion so s o as to make it coincide with the other. other. y a rigid motion motion we mean a transation transation or shift a rotation in the pane p ane or a reection in a ine . he atter can be viewed informa informay y ifting the gure gure from from the pane ipping it over and pacing it back in the pane pan e as ifting In Figure 1.2, for exampe the three trianges are congruent athough to make Z coinc co incide ide with either of the other two triang trianges es it is neces nec essary sary to ree reect ct it or ip ip it over. We write ABC �RST to report that that the rst two trianges in Figure 1 are congruent congruen t but note that there is more to thi thiss notation than may at rst be apparent. he ony way that these two trianges can be made to coincide is i s for point R to coincide with point A , for S to coincide with B , and for T to coincide with C We say that A and R B and S, and C and T are corrpondin point of these these two congruent trianges trianges.. he ony
6
CHAPTER
THE BASCS z R s
T
Figure 1.2
correc correctt ways to report report the congruence is to ist corresponding points in corresponding positions posi tions It is correct therefore therefore to write ABC RST or BAC SRT but it is wrong to write ABC S RT his atter atter assertion is not true because there is no way that that these two trianges trianges in Figure 1 2 can be made to coincide with A and S, B, and R , and C and T being corresponding points points Since RST Z, it is cear cear that corresponding corresponding sides of thes thesee trianges have equa ength and that corresponding correspond ing anges have equa measure (contain (con tain equa numbers numbers of degrees or radians) radians ) We can thus write for exampe R S and LSRT L Z Note that in this context, co ntext, the notation L S RT refers refers to the measure o f the ange in some convenient units units such as degrees or radians radians In other other situations situations however we may write write S R T to refer refer to the the ange ang e itsef its ef his hi s is entirey entirey anaogous anaog ous to t o the fact fact that R S can ca n refer either to a ine segment or to its ength in centimeters inches mies or whatever In some geometry books books the notation mLSRT is used us ed to denote the the measure measu re of LSRT Whenever Whenever we know that two trianges trianges are congruent congruent we can deduce six equaities three three of engths and three three of measures of anges anges It is aso as o reasonaby ovious that given given two trianges trianges,, if a six equaities hod, then the the trianges trianges can be made to t" one on top of the other other and they are congruent con gruent s readers readers of this book boo k are surey aware it is not necessary neces sary to know a six equaities to concude that two trianges trianges ae a e congruent If we ow, for exampe that the three sides side s of one triange equa respectivey, the three corresponding sides of the other triange we can safey deduce that the trianges are congruent cong ruent If we know, know, for exampe that A B RS, AC RT, and B C ST, we can concude that ABC RST In a proof proof where where each assertion ass ertion must be justied ju stied we say that these two trianges are congruent by SS S S S " he abbre abbreviation viation SS S S S which stands for sideside side sideside side"" refers refers to the theorem that says that if the the three sides of a triange triange are equa in ength to the three corresponding corresponding sides of o f some other triange triange then the two trianges trianges must be congruent co ngruent 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 these are entirey famiiar to readers of this book we iustrate ony one of them with an exampe In Figure 12 if we somehow know that S T Z and that L S L and L R L, we can write in a proof We concude by S that SRT Z. (Notice (Notic e that we we wrote as a short form form for for LRST, which is the fu fu name of this this ange ange his is acceptabe accepta be when it cannot resut in ambiguity) ambiguity ) What What is the ogica status of the four congruence criteria criteria S S S SS S S S, S , and S? For each of these these the fact that the criterion criterion is sufcient sufcient to guarant gu arantee ee the congruence of of
B CONGRUENT TRANGLES
7
two triagles triagles is actually a theorem, proved proved by Euclid from from his postlat pos tlates es These The se four four theorems are amog amo g the the basic bas ic results resul ts that we are are acceptig as kow ko w to be valid ad that we are willig to use without providig proofs proofs I fact, however, however, it is ot hard to prove prove the sufciecy of some s ome of these criteria if we are willig to accept some of o f the others s a example, example , ad as the rst proof proof that we we actually preset i this book, bo ok, let us deduce the sufciecy sufciecy of the SSS S SS criterio, criterio, with the uderstadig uderstadig that we may freely freely use ay of the the other three triaglecogruece triaglecogruece coditios ssume that i ABC ad R, we kow that AB R, A C R , a ad BC rov rovee tha thatt A B C R without without usig the the SSS SS S cogruece critero critero
(1.1) PROBLEM.
To do this, this , we w e shall shal l appeal to aother theorem that that readers readers surely s urely remember from from thei previous study of geometry: The base agles of a isosceles triagle are equal ecall that a triagle U V W is ioc if two of its sides have equal legths I Figure 1.3 for example, example, the triagle triagle is isosceles isos celes because U V UW. The third side V W occ urs at the bottom of the the is called the ba of the triagle, whether or ot it actually occurs diagram The ba an of a a isosc is osceles eles triagle are the the two agles at the eds of the the base, base , ad the theorem theorem asserts that they they are ecessarily equal equal I Figure 1.3 therefore, we have U V W U W V. We metio that this baseagles theorem for isosceles triagles triagles is used so s o ofte ofte that it is give a ame: the pon ainorum This ati phrase meas bridge of asses " pparetly, pparetly, the theorem theorem has acquired this this ame partly because the diagram i The Elements used i its proof vaguely resembles a bridge bridge
Figre 13
y reamig the poits, poits , if ecessar ecess ary, y, we ca ass assume ume that that AC is the logest side of ABC, ad cosequetly, R is the logest side of R Sice Si ce we we are are give give that that AC R , we ca mov movee R , ippig it over over,, if ecessary, so that poits A ad R coicide, as do poits C ad , ad so that lie A C poits B ad lie o opposite sides of lie Now draw lie segmet segm et B What must result is a situatio si tuatio resembli res emblig g the left left diagram i Figure 14. It is ot possibl pos siblee for B to fail to meet A C , as i the ight F igure 1.4 or for B to go through oe of the poits A or C because diagram of Figure that would would require require oe of BC or BA to be loger tha tha A C (We (We are are sham s hamelessly elessly relyig o the diagram to see this ) We ca thus assume as sume that we are i the situatio of the left left gure g ure Sice A B R , we see that A B is isos is oscel celes es with base B , ad hece hec e by the pos asiorum, asi orum, we deduce that y , where, as idicated idic ated i the diagram, diagram, we are writig ad y to represet represet the measures measures of A A B ad R B , respectivel respectivelyy Simlarly, Simlarly, by a secod s ecod applicatio of the the pos asiorum, as iorum, we obtai u v ad it
oution to Probm 1.1.
8
CHAPTER
THE BASCS
A
C
R
T
T
s
s
Figre 1.4
follows by additio that that x + u y + v I other words, LABC R Sice Sice we already kow that A B R ad B C , we ca coclude coclude by SS that ABC R , as requ require ired d • Observe Ob serve that we have writte writte this proof i a coversatioal coversa tioal sty e, usi u sig g complete comp lete Eglish seteces s eteces grouped ito logical paragraphs paragraphs We eded by establishig establis hig what we set out to prove, ad we clearly marked the the ed of the the proof (It has become customar cu stomaryy for a box box to replace the oldfashioed oldfash ioed E as a edofproof marker) marker) This style s tyle,, with mior variatios, has become the accepted model for what a proof should look like throughout throughout most mos t areas of mathemat mathematics ics,, ad we follow it cosistetly cosi stetly i this book We expect studets to do the homework exercises with proofs writte i the same style, usig usi g complete seteces s eteces We do ot recommed the the twocolum proof format ormat that is ofte ofte required i highschool high school geometry geometry classes classe s Note that the secod setece of the last paragraph of the proof begis with the word similarly This Thi s word ca be a powerful tool for for simplif simpli fyig yi g ad shortei sh orteigg proofs proofs ad makig them them more itelligible itelligibl e ike most powerful powerful tools, however, however, this oe ca be dagerous dagero us if use improperly improp erly We ecourage ecour age studets to use the word similarly to avoid uecess uece ssar aryy repetitio i their proofs, proofs, but to use it carefully carefully lthough we expect that most readers of this this book remember how to prove the pos asiom, as iom, we preset more tha oe proof here to illustrate a few poits poits ad to provide provide furt further her models of proofwri proofwritig tig style styl e Our rst proof also als o yields some additioal iformatio iformatio about is osceles osc eles triagles triagles We remid the reader reader that a mdian of a triagl triaglee is the lie segmet j oiig a vertex to the the mdpoit of the opposite side s ide THEOREM et A B C be isosc isoscele eless with base BC Then L B L C Also Also the median from ve v e rtex A the bisec bis ector tor of of L A and the altitud a ltitudee from vertex v ertex A a re all a ll the same line
I Figure 1.5 we have draw the bisect bis ector or A of LA, ad thus LBA LCA y hypothesi hypothesis,s, we kow that that AB AC ad of course, A A Thus BA CA by SS S S It follows follows that L B L C sice these are correspodig agles i the coget triagles triagles We also eed to show that the agle bisector A is a media medi a ad that it is a altitude too To see s ee that it is a media, it sufces sufces to check that is the midpoit
Proof.
B CONGRUENT TRANGLES
of segmet B C, ad this is true because B X X C sice sic e these are correspod ig sides of our cogruet triagles Fi ally, to prove prove that AX is also a altitude, altitude, we must show that AX is perpedicular to BC I other words, we eed to estab lish li sh that B X A 90° . From the cogruet triagles, we kow that that the the correspodig agles agl es B X A ad C X A are equal, equal , ad thus BXA BXC 90° sice the the straight agle B X C 180°. •
9
A
B
C
Figre 1.5
To prove that the agle ag le bisector, media, ad a d altitude fro from m vertex vertex A are all the same, s ame, we started started by drawig oe of these lies (the bisector) ad showig that it was aso as o a media ad a altitude Sice there is oly oe media ad oe altitude from A, we ko ko that the bisector bi sector is the media media ad the altitude What would have happeed happee d if we had started by by drawig the media from A i istead stead of the bisector bis ector of A A ? This is, is , after after all, the same lie I this situatio, we could deduce that that B AX C AX by S SS SS It would would the follow ollow that that B AX C AX AX We We would deduce that AX is the agle bisector, ad everythig would proceed as before This appoach would have bee less s atisfacto atisfactory, ry, because it makes the pos asiorum deped o the SSS SS S cogruece co gruece criterio criterio I roblem 1 . 1 , however, however, we proved proved the the validity of of the S S S criterio usig the pos asiorum, ad this this would be a example of ivalid circular reasoig What if we had started by drawig alti A tude AX ? We We would the the kow kow that that B XA C X A sice si ce both of o f these are right agles agle s We also kow kow that that AB AC ad AX AX t this poit, we mght be tempted tempted to coclude that BAX C A X by SS, SS , but we we would would resist B D that temp temptat tatio, io, of course, course, because SS is ot a valid cogruece co gruece criterio To see why, cosider co sider Figre 1.6 Figure Figure 1 .. . . I this gure, the base base B C of isosce les ABC has bee exteded to a arbitrary poit D beyod C The two triagles ADC ad AD B are clearly clearly ot cogue coguett because because DB D C, ad yet the triagles triagles agree agree sidesideaglee sice AB A B AC, AD AD, ad ad D D i sidesideagl There is oe case where SS is a valid cogruece criterio: whe the agle is hypoteu sear arm" m" criterio, criterio, abbreviated abbreviated eca ecallll tha the a right agle This is the hypoteuse logest loges t side of a right right triagle triagle,, the side opposite oppo site the right right agle, is i s called the hypotnu of the triag triagle le The other two sides si des of the triagle are ofte ofte called its arm
two right triangles tri angles have equal eq ual hypoten hypotenuse usess and an arm a rm of of one of the triangles equals equ als an arm of of the othe othe then the n the triangles tri angles are congrue co ngruent nt
( 1 . 3) THEOREM.
10
CHAPTER
THE BASCS
B Figre 1.7
We are give triagles ABC ad D with right agles at C ad We kowthatAB = DadAC = D,adwewatto showthatABC D Move D , ippig it over over ifif ecessary eces sary,, so that poits A ad D ad poits C ad coic co icide ide ad the diagram diagram res resembles embles Figure 17 I Figure 17 we have B C = B C A + D = 90° + 90° = 180° ad thus B C is a lie segmet, which we ca ow call B Sice Sic e A B = D , we see that A B is isosceles isos celes with base base B Thus altitu altitude de AC is a media by Theor Theorem em 1 ad hece BC B C = The desired cogece cogece ow o w follows follows by S SS •
Proof.
We ed this sectio s ectio with with yet aother proof proof of the pos asiorum This Thi s oe o e is amusig ad very short, but it is somewhat tricky tricky ad should sho uld be read carefully carefully We are give give isosceles iso sceles A B C with with base bas e B C , ad we wat wat Sice also also A = A, to show show that that B = C We have AB = AC ad AC = AB Sice we ca coclude that ABC A CB by SS It follo follows ws that that B C sic sicee these are corres correspodig podig agles of the cogruet triagles triagles •
Proof of pon ainorum.
xercses B
rove the coverse of the pos asiorum Show, i other other words, that if B = C rove i A B C, th the AB = AC If the altitude from from vertex vertex A i A B C is also the bisector bise ctor of A , show s how that that A B = AC altitude from from ver verte texx A i A A B C is i s also a media, show that AB = AC If the altitude This fact fact will be used later, later, s o please do this this problem I Figure 18 medias B Y ad CZ have bee bee draw draw i isosceles isos celes A B C with with basee B C Show bas S how that that B Y = CZ assume me ow that BY B Y ad C Z are agle bisectors bis ectors of sig Figure 18 agai, assu isosceles isosceles A B C wit with base base B C Show Show tha thatt B Y = CZ gai i Figure 18 assume that that A B C is isosceles isos celes with with base BC, B C, but this this time, assume tha B Y ad C Z are alti altitudes tudes Show that that B Y = C Z s ssume sume for for this problem that, that, as i the gure, gure, the two altitudes actually actually lie iside isi de the triagle triagle but observe that this does ot always happe happe
ANGLES AND PARALLEL LNES
11
A
Figure 1.8
Now i Figure 1. assume that BY ad CZ are equal altitudes Show that A B AC eferrig eferrig agai aga i to Figure 1. let P b e the the poit where where B Y ad C Z meet ssume that that B Y CZ a ad P Y PZ Show that that A B AC Oce more i Figure Figure 1. let P b e the the itersectio itersectio poit o f BY ad C Z ssume s sume that A B A C ad B Y C Z We ask if it is ecessarily eces sarily te that P Y P Z ou are asked, i other words, either to prove prove that P Y P Z or else to show show how to draw a couterexample couterexample diagram diagram where all of the hypotheses hypothes es hold but where P Y ad P Z clearly c learly have differet differet legths legths We drew drew Figure 1 so that P Y ad P Z actually are equal, ad so if a couterexample couterexample exists, exists , it would ecessarily eces sarily have to look somewhat so mewhat dieret from from the gure
C
Angles Angl es and Parallel Lines
tranvra is a lie that cuts across two give give lies sually, sually, the two give lies are are parallel whe we use the word transversal but we do ot ot absolutely isis i sistt o this I Figure 1.9 for for example, lie li e t is a trasversal trasversal to lies a ad b There There is some omeclature that that is usefu us efull for for discussig discus sig the eight agles a gles that we ave l abeled with 1 through i Figure 1.9 gles that are are o the same side of the trasversal trasversal ad o corresp correspodig odig sides of the two lies a ad b are said to be corrpondin ales I Figure 19 therefore, L 1 ad L 5 are correspodig agles, as are L ad L ad also L3 ad L 7 ad, of course, L 4 ad L. It is a theorem theorem that that correspodig agles agl es are are equal whe a trasversal cuts a pair of parallel lies, ad coversely, coversely, if i Figure 19 ay ay oe oe
a b
Figure 1.9
12
CHAPTER
THE BASCS
of the equalities L 1 L5 L L L3 L7 or L4 L 8 is kow to hold, the it is a theorem theorem that lies a ad b are parallel, parallel, ad thus thus the other three three equalities also als o hold airs of agles such as L4 ad L or L 3 ad L5 that lie o opposite oppos ite sides of the trasversal ad betwee the two give lies are called atrnat intrior agles ad pairs such as L 1 ad L 7 or L ad L 8 that lie o opposite sides s ides of o f the the trasversal ad outside of the space betwee the two parallel lies are atat xtrior agles It is a theorem that alteate iterior agles are equal ad that alterate exterior agles are equal whe a trasversal cuts two parallel lies It is also als o true that, coversely c oversely,, if i Figure 19 ay oe of the equalit equalities ies L 1 L7 L L8 L3 L5 or L4 L is kow to hold, the lies a ad b must be parallel, ad thus the other three three equalities also hold, as do the four four equalities eq ualities metioed me tioed i the previous paragraph paragraph We recall also that whe two lies cross, as do lies a ad t i Figure 19 the agles , of o f course, L 1 ad L 3 are said to be vrtica an, as are L ad L4 ertical agles, are always equal While reviewig reviewig omeclatur omecl aturee for for agles, agles , we metio that two agles whose measures sum to 180° are said to be uppmntary, ad if the sum is i s 90° the an, ad a agle ages age s are are compmntary Of course, a agle of 180° is a traiht an, of 90° is a riht an I Figure 19 we see se e that L 1 ad L4 are supplemetary, ad so if a ad b are parallel, parallel, the L 1 L5 ad it follow followss that L4 ad L 5 are supplemetary, suppl emetary, as are L 3 ad L We ca apply some s ome of this to the agles ag les of a triagle Give Give ABC exted side B C LAC D is said to poit D as show i Figure 110 I this situatio, LACD said to be a xtrior agle of the triagle at vertex C, ad the two agles L A ad L B are the rmot iterior agles with respect to this exterior agle THEOREM. An exterior angle of a triangle equals the sum of the two remote interior inte rior angles Also Also the sum of of all three interior interio r angles angles of of a triangle is 1 8 0° 0°
Figre 1.10
We eed to show i Figure 110 that LACD LA + LB. raw a lie CP through C ad parallel to AB as show show Now L A LACP sice sic e these are alteate alteate iterior agles for parallel parallel lies li es A B ad P C with respect to the trasversal A C lso, L B L PC D sice these thes e are correspodig correspodig agles It follows follows that LACD LAC D required ed LACP LACP + L P C D LA + B , as requir LAC D+ LAC B L B CD C D 180° The substitutio We see se e i Figure 110 that LACD+ of L A + L B for L AC D i this equatio yields the desired coclusio coclu sio that the sum of the the three iterior agles of ABC is 180° •
Proof.
Observe that our proof that the iterior agles of a triagle total a straight agle relies o the fact fact that it is pos sible to draw a lie through C parallel to A B I fact,
I ANGLES AND PALLEL LNES LNES
1
there does ot exist a proof of this result that does ot, omehow, deped o paallel lies This is because i oEuclidea geometries, it is ot true that the agles of a triagle must total 1 80 80 , ad yet the oly fudametal fudametal diff differece betwee betwee Euclidea Euclid ea ad ad oEuclidea oEucli dea geometries is i the parallel postulate triagle, of course, is a polygo with three sides We digress to cosider the questio of how to d the sum of the iterior iterior agles agles of a polygo with w ith sides, sides , where where may be larger tha thee Fid a formula formula for the sum of the iterior iterior agles of a go go osider, o sider, for for example, the case 6 I Figur Figuree 1 1 1 , we see two two 6gos, which which are usually usu ally called hexagos hexago s I the left left hexago, we have draw the three diagoals diagoals from vertex A ( diaona of a polygo is a lie segmet joiig two of its oadj oadj acet vert vertices ices ) I geeral, a go has exactly exactly - 3 diagoals diagoal s termiatig termiatig at each of its vertices vertices,, ad this gives a total total of ( - 3 /2 diagoals i all all (o you see s ee why we had to divide divid e by 2 here?) polygo is convx if all of its diagoals lie etirely i the iterior The iterior agles of a polygo are the agles as see from iside, ad for a covex covex polygo such as the left left hexago i Figure Figure 1 1 1 , these agles are all less tha tha 1 80
A
D
Figre 1.11
I the right hexago o f Figure Figure 1 1 1 , which is ot covex, covex, we see s ee that that two two of its six iterior agles exceed 180 ( agle with measure lager tha 180 is sai to be a rx agle ) I geeral, a polygo is covex covex if ad oly if its iterior iterior agles all measure meas ure lss l ss tha 1 80 If som itrior agle agle of a go g o is exactly exactly equal w e do ot cosider it to be a covex go, go, but i this situatio, si tuatio, there are are to 1 80, we s ides that together together form form a sigle lie segmet, ad the go go ca c a be two adjacet sides cosider cosidered ed to to be a a ( 1) go iewed iewed as a ( 1 )go, the give polygo polygo may may be covex Suppos Suppo s w have have a cov covxx go go such as hxago AB A B CD F of Figu Figurre 1 1 1 Fix some som e particular vertx vertx A ad draw the 3 diagoals from from A A This divides the origal polyg ito it o exctly 2 triagles triagles ad it should be clear that the sum of all th trior agls of all of ths triagls is xactly the sum of all iterior agles of th origial polygo po lygo It follows follows that th sum of the iterior iterior agles of a covex covex polygo is xactly xactly 1 80( 80 ( 2) dgrs dgrs W have ot yt yt fully fully solve so lve roble roblem m 1 5 , of course, because becaus e we have have oly cosiderd cosi derd covex covex polygos polygo s ctually, it is ot quite ecessary for for the polygo poly go to b covx to mak th prvious argumt work What is really required is that
14
CHAPTER CHAPTER
THE BAS CS
there should exist exi st at least oe vertex vertex from from which all of the diagoals lie iside i side the polygo polygo ecall that the deitio of a covex covex polygo requires requires that all diagoals from all vertices should be iterior It is coceivable that every polygo has at least oe vertex from which all diagoals are iterior, but ufortuately, that is ot true true car c aref eful ul ispectio of the right right hexago hexago i Figure Figure 1 1 1 shows that it is a couterexample; at least oe diagoal diago al from from every every oe of its vertices fails to be iterior iterior It is true, true, but ot easy to prove, that every every polygo has at least oe iterior diagoal, diago al, ad it is possibl pos siblee to use this hard hard theorem theorem to costruct co struct a proof that that for for ever everyy go go, , the the sum of the the iterior agles is 1 80( 80 ( 2) degrees There is aother way to thik about roblem 15 that may give additoal isight isi ght Imagie walkig clockwise clockwi se aroud a covex covex polygo, startig from from some poit other tha a vertex o oe of the sides Each time you reach a vertex, you must tu tu right by a certai umber of degrees degrees If the iterior agle at the th vertex vertex is fk the it is easy e asy to see that the the right tu at that vertex vertex is a tu through precisely preci sely 1 80 - fk degrees Whe you retu to your startig poit, you will be facig i the same s ame directio as whe whe you started, started, ad a d it should be clear that you have tured clockwise clockwi se through a total total of exactly exactly 360 360 • I other words, words,
L
(1 80 - f L k=
360
L
Sice Si ce the quatity 1 80 is added times times i this sum ad each each quatity f is subtracted oce, we see that that 1 80 - fk 3 60, ad hece hece fk 1 80 80 - 360 180( - 2) This provides provid es a seco s ecodd proof of the formula for the sum of o f the iterior iterio r agles of a covex covex go ow importat is covexity for this secod s ecod argumet? argumet ? If whe walkig wal kig clock wise aroud the polygo pol ygo you yo u reach the th vertex vertex ad see a reex iterior agle there, you actually tu left, ad ot right I this case, your left left tu is easily see to be through exactly exactl y f - 1 80 degr degrees ees,, where, where, as before before,, f is the iterior iter ior agle at the vertex If we we view vie w a left tu as beig bei g a right tu through some so me egative e gative umber of degrees, degrees , we see s ee that at at the th vertex vertex we are are turig right by 1 80 - fk degrees, ad this is i s true regardless regardless of whether whether fk < 1 80 as i the covex covex case or f 1 80 at a reexagle vertex vertex It is also als o clearly true true at a straightagle straightagle vertex, where fk 180 The pre previous vious calculatio thus works works i all cases, ad it shows shows that 1 80( 80 ( - 2) is the sum of o f the iterior agles for every every polygo, polygo , covex covex or ot ot
ID
Parallelograms
mog polygos, perhaps parallelograms are secod i importace after triagles ecall that, that, by deitio, dei tio, a paraoram is a quadril quadrilate ateral ral AB C D for for which which A B CD ad A D B C I other words, the opposite oppos ite sides sid es of the the quadrilateral quadrilateral are parallel It is also true that the opposite oppos ite sides si des of a parallelogra parallelogram m are equal, but this is i s a cos c osequece equece of te te assumptio assumptio that the the opposite oppos ite sides are parallel; it is ot part of the deitio deitio (1.6)
pposi te sides s ides of of a parallelogram pa rallelogram are a re equal equ al Opposite
D PARALLELOGRAMS
15
I Figure 1 1 2, we are give that that A B C D ad AD A D B C, ad our task task is to show thatAB CDadAD BC raw diagoal BD ad ote ote that A BD CDB sice these these are are alterat alteratee iterior iterior agles for for the paralle parallell lies A B ad C D , ad simila similarly, rly, D BC A DB Sic Sicee BD B D, we see see tha thatt D AB B CD by by • S, ad it follows follows that that AB A B CD ad ad AD B C, as require required d
Proof.
B
C
D
Figre 1.12
Figre 1.13
There are are two useful coverses for Theorem Theorem 1 6 In quadril quadrilate aterral A B C D sup suppose that that A B CDandAD BC ABC D is a parallelogram. Then ABCD In quadrilateal quadrilate al ABCD suppose suppose that A B C D and ABCD Then ABCD ABC D is a parallelogram.
We leave the proofs of these thes e two theorems to the exercises exercis es A quadrilateral is a parallelogram and only its diagonals bisect each each othe
s show i Figur Figuree 1 1 3 , we let X be be the the poi poitt where where diagoa diagoalsls AC ad B D of quadrilat quadrilatera erall A B C D cross Suppose rst that that X i s the commo commo midpoit of of lie lie segmet segmetss A C ad BD The The AX XC ad BX XD ad ad aalso lso A XB CXD because these are are vertica verticall agles It follows that that AX B C X D by b y SS S S , ad thus AB CD Similarly, AD BC, ad hece ABCD ABCD is a parallelogrm parallelogrm by Theorem 17 oversely oversely ow, ow, we assume that that AB A B C D is a parallelogram, ad we show sh ow that that X is the midpoit of each of the diagoals A C ad B D We have have B A X X C D ad ABX AB X CDX CD X because bec ause i each each case, cas e, these are pairs pairs of alteate alteate itrior itrior agles agles for the the parallel parallel lies lies A B ad CD C D lso, A B C D by Theorem 1 6, 6 , ad thus ABX C DX by S We deduce deduce that that AX CX ad BX DX, as • required
Proof.
Observe that that by Theorem Theorem 1 7, 7 , a quadrilateral quadrilateral i which all four four sides are are equal must be a parallelogram parallelogram Such a gure g ure is called a rhombu I the case of o f a rhombus the diagoals ot ot oly bisect bise ct each other, but they they are als alsoo perpedicular perpedicular lthough lthou gh this fact fact is ot difcult to prove directly, we prefer prefer to derive it as a coseque cos equece ce of the follo followig wig more geeral result
16
CHAPTER
THE BASCS
iven a line segment BC BC the locus of of all points po ints equidistant eq uidistant from B and C is the perpendi perpendicular cular bisector bise ctor of of the segment
(1.10)
We eed to show that every poit o the perpedicular bisector of BC is equidistat equidistat from from B ad C ad we must also show that every every poit that is equidistat from rom B ad C lies o the perpedicular bisetor of B C ssume ss ume that A X is the perpedic perpedicular ular bisector bisector of o f B C i Figur Figuree 1 14 14 This This meas meas that X is the midpoit of B C ad AX is perpedicular to to BC I other words, we are assumig that A X is simultaeously s imultaeously a media ad a altitude altitude i ABC ad we wat to deduce that A B AC This is precisely precisely Exercise Exercise 1 3 , ad we will will ot give the proof here ssu s sume me ow that A is equidistat from B ad C i Figure Figure 1 1 4 ad draw draw meda meda A X of ABC Sice A B AC this triagl triaglee is isosceles, isosc eles, ad thus by Theorem Theorem 1 2, 2 , media A X is also a lso a a altitude altitude I I other other words, words, A X is the perpedicular bisector of B C ad of course, A lies o this lie lie •
Proof.
A
B
X
C
Figre 1.14
(1.11)
of rhombus A B C D are per pe rpendicu pe ndicula la The diagonals of
Sice A B AD we kow by Theorem Theorem 1 1 0 that that A lies o the perpedicular bisector bis ector of diagoal B D ad similarly, C lis o the perpedicular bisectr of B D ut A C is the oly li that cotais the two poits A ad C ad thus A C is the perpedicular bisector of B D. I particular, diagoal A C is prpeicular to diagoal B D •
Proof.
We close clo se this sectio sectio by metioig aother spcial typ typ of paralllogram paralllogram a rct rct agle y deitio, a rctan is a quadrilateral quadrilateral all of whos whos agls agl s r right agles agles It is easy it see that th opposite sies s ies of a rctagl rctagl parallel parallel d so a rctagle is automatically automatically a parallelogram W also kow (by roblm 1 , fr fr xampl) tha th sum of th four four agls agl s of a a arbitrary arbitrary quarilatral is 0 0 hus if i f w hav ay quadri later la teral al with all four agls qual each agl must b 90 ad th gur is a rctagl rctagl W rmark also that adjact vrtics of a parllelogram hav supplmt itrior agles agles It follows easily that if o gle of a paralllogram paralllogram is a right agl ag l th th parallogram must mu st b a rctagl rctagl Fially w mtio that a quadrilatrl that is both a rectagle ad a rhombus rhombus is by itio a ar
ID PARALLELOGRAMS
xercses
17
rove rove Theorem Theorem 1 7 by showig sho wig that that quadrila quadrilatera terall A B C D is a parallelogram if A B CD ad ad AD BC rove Theorem 1 8 by showig that that quadrilat quadrilateral eral AB A B C D is i s a parallelogram if A B CD a ad AB CD rove that the diagoal diag oalss of a rectagle are equal equal rove that a parallelogram paralle logram havig perpedicular diagoals diagoal s is a rhombus rhombus rove that a parallelogram parallelogram with equal diagoals is a rectagle rectagle I Figur iguree 1 15, weare weare giv give that AB CD ad that A D ad B C are ot parallel paralle l Show Sh ow that D C if ad oly if A D BC HINT raw a lie through B parallel Figure 1.15 toAD NOTE: ecall ec all that a quadrilateral is said to be a trapzoid if it has exactly oe o e pair of parallel sides si des If the oparallel pair pair of sides are equal, the trapezoid is said to be ioc Show that opposite agles of a parallelogram parallelogram are equal equal I quadrila quadrilater teral al AB A B C D, suppose that that AB C D ad B D Show that that AB C D is a parallelogram I quadr quadrililat ater eral al AB CD , supp suppose ose that A C a d B D (We are are refer eferrrig to the iter iterior ior agles, agles, of course) course ) Show Sh ow that that AB C D is a arallelogram HINT: Show that A ad C are supplemeta supplemetary ry that the diagoals of a iso sceles scele s trapezoid trapezoid are equal Show that I Figure Figure 1 6,we are are give give that that �AX �A X B ad �A Y B are are isosceles isosc eles ad share share base S how that poits X, Y, ad Z are are colliear (lie o a commo lie) if ad ad A B Show oly if �A ZB is isosceles wit withh base base AB AB x
Figure 1.16
18
CHAPTER
THE BASCS
I Figure 117, poit is equidis tat from the the vertice verticess of �A B C Thi Thi s poit is reected i each side of the triagle, yeldig poits X, Y, ad Z Z This meas that X is perpedicu lar to B C ad that the perpedicular perpedic ular C distace dis tace from X to B C isi s equal equ al to that from to BC Similar asser x tios hold for Y ad Z rove that �ABC �XYZ ad that corre Figure 1.17 spodig sides side s of these two triagles triagles are parallel HINT: Show that quadrilaterals ZBA ad CAY are rhombuses educe that B Z ad C Y are are equal ad parallel NOTE Give ay triagle, there always exi sts a poit equidistat equidi stat from from the three three vertices vertices This poit, called the circumcntr of the triagle, triagle, does ot always lie iside isid e the triagle The assertio as sertio of this problem is still s till valid if lies outside of �A B C or eve if it lies lie s o oe of the sides side s Each pair of parallel lies i Figure 118 represets a railroad track with with two parallel rails The distaces betwee the rails rail s i each of the tracks tracks are equal rove rove that the parallelogram formed formed where where the tracks tracks cross cros s is is a rhombus rhombus Figure 1.18 NOTE The distace betwee the rails is, of course, measured perpedicular perpedicul arly ly It shou s hould ld be clear that for for each track, the perpedicular perpedic ular distace dista ce from from a poit o oe rail to the other rail is costat, cos tat, idepedet idepe det of the poit dis tace from from a poit to a lie is measured perpedicularly Give ecall that the distace A B C, show that the locus locu s of all all poits iside i side the agle ad equidistat equidis tat from from the two lies B A ad a d B C is i s the bisector bis ector of the agle
IE
Area
We assume ass ume as a s kow the t he fact fact that the area of a rectagle is i s the product of its legth ad width The area of a geometric gure g ure is ofte deoted K , ad so the formula for the area of a rectagle is K = bh where b is the legth of oe of the sides si des of o f the rectagle ad h is the legth of the perpedicular sides side si de of legth legth b is referred referred to as the base b ase of the rectagle, rectagle, ad the height h is the legth of the sides perpedicular to the base Of course, the base of a rectagle rectagle does ot have to be at the the bottom; it ca be ay side si de Now suppose that we have a parallelogram that is ot ecessarily a rectagle ga we desigate de sigate oe o e side si de of the the parallelogram parallelogram as the base ad write b to deote its legth ut ut this time, the height h is the perpedicular distace betwee the two parallel
E AREA AREA
x
I I hi I b
19
b
x
b Figure 1.19
b
sides of legth Of course, if the parallelogram parallelogram happes to be a rectagle rectagle,, the this perpedicular distace is the legth of the the other other two sides side s O the left of Figure 119, we see a rectagle with base ad height h . O the right, we have draw a parallelogram that also has base ad height h ad we claim areas To see why this is true with that the parallelogram ad the rectagle have equal areas a very iformal argumet, drop perpediculars from oe ed of each of the sides of legth of the the parallelogram parallelogram to the extesio extesios s of the the opposite sides s ides,, as a s show sho w i the gure What results result s is a rectagle with base base + x ad height h where, as show, x represets represets the amout that the the bas basee of the parallelogram parallelogram had to be b e exteded to meet the perpe perpedicu dicular lar The area of this rectagle rect agle is + x )h ad to obtai obtai the area o the origial parallelo para llelogram, gram, we eed to subtract from from this the area of the two right triagles learly, learly, the two tw o right triagles triagles pasted together would form form a rectagle with bas e x ad height h ad so the total area of the two triagles triagle s is x h The area of the parallelogram parallelo gram is therefore + x)h - xh h , ad so the formula K h works to d areas of arbir arbirary ary parallelograms ad ot just of rectagles rectagles other way to see why the the rectagle ad ad parallelogram parallelogram i Figure 1 1 9 have equal area is from the poit of view of calculus Imagie slicig each of the areas ito iitesimally iites imally thi horizotal strips of legth We see se e that each of the the two areas is is give by the same s ame formula formula
b b
b
b
(b
(b
b
b
b. K lhbdY,
equal , as claimed cl aimed I fact, fact, if we carry carry out the itegratio, itegra tio, we we ad thus the two areas are equal, expect get the th e formula formula K h , as we expect Next, we cosider co sider areas of triagle triagless I the left left diagram diagram of Figure 1 20, w have draw draw a triagle with base ba se ad height h where ay ay oe of the three three sides ca c a be viewed as the base ad the legth le gth of the altitude draw to that that side is the correspo correspodig dig heght heg ht emember emember that this this altitude altitude may lie l ie outside of the triagle triagle
b
b
B
Figure 1.20
X
C
20
CHAPTER
THE BAS CS
To compute comput e the area K of the the give triagle, we have costructed co structed a parallelogram parallelogram by drawig lies parallel to our base ad to oe of the other sides of the the triagle, as as show i the gure This parallelogram has base b ad height h , ad so its area is bh The parallelogram parallel ogram is divided divid ed ito two cogruet triagles by a diagoal diago al,, ad so the area of each of these thes e tria triagle gless is exactly exac tly half the area of the parallelogram (The two tria triagles gles are easily easily see to be cogr co gruet uet by by SSS S SS)) Sice the origial origial triagle triagle is i s oe of the two halves of the the parallelogram, we see that K bh ad this is i s the basic formula formula for for the area of a triagl triaglee s is idicated idi cated with A A B C o the right of Figure 1 .20, .20 , ay oe of the three sides coud have bee desigated de sigated as the base, bas e, ad for for each, there is a correspo correspodig dig altitude (It i ot a coicidece coi cidece that the the three three altitudes are cocurret; cocurret; we shall sh all see i hapt h apter er 2 that thi thiss is guarateed to happe) happe ) We thus thus hav h avee three dieret formula formulass for the the area area KAB of A B C We have
KAB
AXBC 2
BY·AC 2
CZ·AB 2
ad we ca deduce ded uce usefu u sefull iformatio iformatio relatig the legths of the sides side s ad altitudes of a triagle triagle For example, i Exercis Exercisee 1 .6, .6 , you were asked asked to show that that i a isosceles iso sceles A B C with base B C, the t he two altitude altitudess B Y ad ad CZ C Z are are equal, ad you were were allowed allowed to asume that the altitudes were iside is ide the triagle triagle y the area foula foula we have just ju st derived, derived, we kow that B Y · A C C Z A B sice each of these these quatities quatitie s equals equal s twice the area of the triagle triagle We ca cacel the equal equal quatities AC A C ad AB A B to obtai B Y C Z, as desired This proof, furt furthermore, hermore, works idepedetly of whether or ot the altitudes lie iside i side the triagle We remark remark als alsoo that the coverse coverse is true: If we are give that that altitude altitudess B Y ad C Z are equal, equal, it follows follows fro from m the the foula foula BY B Y · A C CZ ·A B that that A B A C (This coverse coverse appea appeared red as Exer Exercise cise 1 .7.) .7 .) We ca get some additioal ifoa ifoatio tio about areas of triagles by b y usig usi g a ittle elemetary elemetary trigo trigoometr ometry y I Figure Figure 1 .2 1 , we have have followed followed custom ad used the symbols a , b ad to deote the the legths of the sides of A B C opposite vertices vertices A, B, ad C, respectively respectively lso, lso , w e have draw draw the altitude of legth h from A (I the gure, this altitude happes happes to lie outside the triagle, triagle, but this is irrelevat irrelevat to our calculatio) calcul atio) We have, of course, K ah, where, as usual, K KA B represets the area of the triagle triagle We see that that si (C) h j b , ad hece h b si(C) si( C) , ad if we substitute substitute this this ito the area formula K ah, we obtai K ab si(C) si( C) Similar Si milarly, ly, we also also have have ad K b si(A) K a si(B) ad I t is ow clear c lear that for for ay ay triagle, the equatios ab si(C) b si(A) a si(B)
mus always hold I f w e divide tough tough by the quatity quatity abc ad take reciprocals, reciprocals, we obtai obtai the socalled soc alled law of sies : si(C)
a si(A)
b si(B)
IE AREA
21
A A
h
B B
Figure 121
C
a Figure 122
simpl e proof of a more powerf powerful formula formula called the exteded law of o f sies that There is a simple applic atio of the sie formula for the area of a triagl triaglee we prove i hapter 2 other applicatio is the followi followig g
et AX be b e the bisector of LA in 6A 6 A B C. Then BX AB XC AC In oher oh er words words X divides B C into int o pieces pi eces propo proportiona rtiona to the engths e ngths of of the near nea rer sides si des of of the trian t riange. ge. THEOREM.
I the otatio otatio of Figure 1 22, we must show that that u / = /, where where ad c the legths le gths of B X ad XC as show show To see see wh thi thiss are as usual ad u ad are the equatio holds, let h be the height of 6ABC with respect respect to the base BC. The h is also als o the height of each each of 6A B X ad 6A C X with with respect to bases base s B X ad XC respectivel We have h cx si() uh b x si() ad = KABX = = KA x = 2 2 2 2
Proof.
where x = A X ad we have writte = LA. ivisio of the rst of these • equatios equatios b the th e secod seco d ields the th e desired propo proportio rtio s a applicatio, applicatio , we have have the followig followig,, which whic h should shoul d be compared with Exer Exer cises cis es 1 2 ad 1 1 3 like those exerci exercises ses,, it it difcult difcult to see a direct ad ad elemetar elemetar proof via cogruet co gruet triagles for for this fact PROBLEM. Suppose Suppos e that that i 6ABC the media me dia from vertex A ad the bisec tor of L A are the same lie Show that A B = AC. I the otatio of Figure 1 22, we have have u = sice the agle bisector bi sector A X is assumed assu med to be a media It follows follows from from Theorem Theorem 1 1 that that a = , as required required •
oution.
22
CHAPTER CHAPTER
THE BASC S
We have ow give two t wo dieret types of formula formula for the area of a triagle: triagle : oe usig oe side ad a altitud altitudee ad the other other usig two sides ad a agle Sice the SS S cogruece criterio tells u s that a triagle is determed determed by its three sides, sides , w wee might expect that there there should be a ice way to compute the area of a triagl triaglee i terms of the the legths of its sides There is I hapter 2, we prove prove the followig followig formula, formula, which is attribut attributed ed to ero of lexadria (c 50 AD : KA B C
a ) (s - b ) ( s
=
-
c)
,
where s (a + b + c) / 2 is called call ed the miprimtr of the triagle triagle Of course course,, a, b, ad c retai their usual usu al meaigs meaigs We close clo se this sectio with with what what seems to be a amazig fact fact s show i Figure Figure 1 23 , poits P Q, ad a d R lie o the sides of �A B C oit P lies li es oe third of the way from from B to C poit poi t Q lies lie s oe third of of the l ies oe third third of the the way from from A to B ie segmets way from from C to A ad poit R lies A P B Q, ad ad C R subdivide the iterior of the triagle triagle ito three quadrilaterals quadrilaterals ad four triagles triagles,, as sh show, ow, ad we see that exactly oe of the the four four small triagles has o vertex vertex i commo com mo with ABC rove that the the area of this this triagle is i s exactly oe seveth of the area of the orgial triag triagle le
(1.14)
We eed to compute compu te the area Kxyz i Figu Figure re 1 23 y choosig uits appro appro priately, priately, we ca c a assume ass ume that KAB 3 ad we write k to deote KByP We draw lie segmet Y C ad start computig areas Sice �CYP has the same s ame height height as 6BYP but its base P C is twice as log, we deduce that K yp 2k Similarly Si milarly,, sice A Q 2 Q C , we see s ee that KABQ 2 K B Q , ad thus K B Q KA B 1 This eables eables us to compute that that K YQ 1 KBY 1 3. y the usual reasoig, reasoig, KA YQ 2 KYQ 2 - 6k, ad we kow that KABQ KAB 2 It follows that KA B y 2 - KAYQ 2 - (2 6k) 6k) 6k owever, KABP KA B 1, ad thus KByp 1 - 6k ut we ow that KB yP k, ad hece 1 - 6k k ad k 1/7 Simlar reasoig shows that KARX 1/7 K Q z ad sice KA R area of quadrilatera quadrilaterall A X Z Q is 1 2/7 5/7 K AB 1 , we deduce that the area Fially, we recall that KA YQ 2 - 6k 8/7, 8/ 7, ad it follows follows that Kxyz 8/7 5 /7 3/ 7 This is exactly oe seveth of the area area of the origial tragle, tragle, as desired desired •
oution.
A
B
P
Figure 123
C
I CRCLES AND ARCS
23
xercses
raw raw two of the the medias of a triagle triagle This subdivides the iterior of the triagle triagle ito four pieces : three triagles triagle s ad a quadrilatera quadrilaterall Show S how that two of the three three small triagles triagl es have equal area ad that the area area of the third is equal eq ual to that of te quadr lateral arbitrary poit P is chose chose o side B C of AB C ad perpediculars P U ad P V ae draw from P to the other two sides of the trigle (It may be that or ies o a extesio of AB or AC ad ot o the th e actual side of the triagle triagle This ca c a happe, happe, for for istace, if L A is obtuse obtus e ad poit P is very ear B or C.) Show Sh ow that that the sum P U + P V of the the legths leg ths of the two perpediculars is costat as P moves alog B C. I other words, this thi s quatity is idepedet of the choice of P Sice Si ce a triagle i s determied determied by aglesideagle, aglesi deagle, there there should shou ld be a foula foula for for formula KA B C expressed terms of a ad L B ad LC. erive such a formula et P be a arbitrar arbitraryy poit i the iterior iterior of a covex quadrilateral quadrilateral raw the li segmets joiig joii g P to the midpoits of each of the four four sides, side s, thereby thereby subdividig the iteror ito four four quadrilaterals quadrilaterals Now choose choos e two of the small quadrilaterals quadrilaterals ot havig a side i i commo ad show that the the areas of thes thesee two total exactly half the area of the the origial gure
IF
Circes an Arcs
s all readers of this this book bo ok surely ow, a circ is the locus lo cus of all poits equidistat equidistat from some give poit called the cntr The commo distace r from the ceter to the poits of the circle is the radiu, ad the word radius is also used to deote deote ay ay oe of of the lie segmets s egmets joiig jo iig the ceter to a poit of the circle chord is ay lie segmet j oiig two poits of a circle, circle , ad a diamtr is a chord that goes through te ceter The legth d of ay diameter is give by d 2r ad this this is the maximum of the legths chords Fially, we metio that ay ay two circles with equal radii are cogruet, of al chords o e of two two cogruet circles ca c a be made to to correspod to ay poit ad ay poit o oe o the other other circle circle Just as ay a y two poit deermie a uique le, le , it is lso true tat ay three poits, uless ul ess they happe to lie o a lie, lie o a uique circle e e is exacty one circe c irce through through any three three given noncoinear Tee points
all the poits A B ad C. Sice by hypothesis, there there is o lie though though these poits poit s we ca be sure that we are are dealg with three three distict poits poi ts,, ad we s egmets A B ad A C. et et ad ad c be the perpedicular bisectors of these draw lie segmets segets ad obs obser erve ve that that lies ad c caot be parallel parallel because lies A B ad A C are either parallel or are are they the same lie lie (Si ( Sice ce A B ad A C have poit A i
Proof.