EuclideanGeometry
Inthischapterwestartoffwithaverybriefreviewofbasicpropertiesofangles,lines,andparallels.Whenpresentingsuchmaterial,one hastomakeachoice.Onecanpresentthebasicresultsofplanegeometryfromfirstprinciples,startingwithanaxiomaticsystem,such asHilbert’sAxioms,carefullylayingoutsuchconceptsasbetweenness,incidence,congruence,andcontinuity.Thisapproachhasseveral virtues.Studentssee,perhapsforthefirsttimeintheirmathematical careers,alogicalsystembuiltentirelyfromfirstprinciples.Theyalso canclearlydeterminewhattheorems,definitions,andaxiomsarefair gametouseintheirownproofsofresults.Ontheotherhand,athoroughand complete developmentofHilbert’saxiomswouldnecessarily takeasubstantialportionofasemester-longcourseingeometry,leavinglittletimeforother,equallyimportanttopicssuchasnon-euclidean geometryandtransformationalgeometry.
Asecondapproachistoreview,insummaryform,someofthe mostimportantlogicalproblemsofclassicalEuclideangeometrythat axiomwriterssuchasHilbertattemptedtofix,andthentomoveon tomoresubstantialresultsinplanegeometry.Thisistheapproach takeninChapter2.Ithastheadvantageofexposingstudentstothe logicalissuesfacingmathematiciansoverthelastseveralhundredyears and,atthesametime,coveringsignificantgeometricideassuchasthe definitionofarea,cevians,andcircleinversion.Onedisadvantageof thisapproachisthatstudentsmayfeelunsureofwhattheycanassume andnotassumewhenworkingonproofs.IneachsectionofChapter 2theauthortriedtocarefullydescribewhatresultsandassumptions weremadeinthatsection.Forexample,insection2.1,studentsare instructedtousethenotionof betweenness inthewayone’stuition
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Instructor’sGuidefor ExploringGeometry,SecondEdition woulddictate,whileatthesametimepointingoutthatthisisoneof thosegeometricpropertiesthatneedsanaxiomaticbase.
IfthemorerigorousapproachtoEuclideanGeometryisdesired,the completefoundationaldevelopmentcanbefoundinon-linechapters attheauthor’swebsite:http://www.gac.edu/ hvidsten/geom-text.
SOLUTIONSTOEXERCISESINCHAPTER2
2.1Angles,Lines,andParallels
Thissectionisperhapstheleastsatisfyingsectioninthechapterfor students,sincemanytheoremsarereferencedwithoutproof.Itmaybe helpfultoremindstudentsthattheseresultswerenodoubtcovered ingreatdetailintheirhighschoolgeometrycourse,andthatafull developmentofsuchresultswouldentaila“fillingin”ofmanydays (weeks/months)offoundationalworkbasedonHilbert’saxioms.
Asignificantnumberoftheexercisesdealwithparallellines.This isfortworeasons.Firstofall,historicallytherewasagreateffortto proveEuclid’sfifthPostulatebyconvertingitintoalogicallyequivalent statementthatwashopedtobeeasiertoprove.Thus,manyofthe exercisesnicelyechothishistory.Secondly,parallelsandtheparallel postulateareattheheartofoneofthegreatestrevolutionsinmath— thediscoveryofnon-Euclideangeometry.Thissectionforeshadowsthat development,whichiscoveredinChapters7and8.
2.1.1 Ithasalreadybeenshownthat ∠FBG ∼ = ∠DAB .Also,by theverticalangletheorem(Theorem2.3)wehave ∠FBG ∼ = ∠EBA andthus, ∠DAB ∼ = ∠EBA.
Now, ∠DAB and ∠CAB aresupplementary,thusaddtotworight angles.Also, ∠CAB and ∠ABF arecongruentbythefirstpartofthis exercise,astheseanglesarealternateinteriorangles.Thus, ∠DAB and ∠ABF addtotworightangles.
2.1.2 Let∆ABC beatriangle,andconsiderthesumoftheangles at A and B .Extendtheangleat A tocreateanexteriorangle.Then, thesumofthisexteriorangleandtheangleat A is180degrees,asthey makeupaline.However,bytheExteriorAngleTheoremweknowthat theexteriorangleisgreaterthantheangleat B .Thus,thesumofthe anglesat A and B islessthanthesumoftheangleat A anditsexterior angle,whichis180.
2.1.3 a.False,rightanglesaredefinedsolelyintermsofcongruent angles.
b.False,anangleisdefinedas just thetworaysplusthevertex.
EuclideanGeometry 13
c.True.Thisispartofthedefinition.
d.False.Theterm“line”isundefined.
2.1.4 a.Apoint M isthemidpointofsegment AB if M isbetween A and B and AM ∼ = MB .
b.Theperpendicularbisectorofsegment AB isalinethroughthe midpoint M of AB thatisperpendicularto ←→ AB
c.Thetriangledefinedbythreenon-collinearpoints A, B , C isthe unionofthelinesegments AB , AC , BC .
d.Anequilateraltriangleisatrianglewhosesidesarecongruent.
2.1.5 PropositionI-23statesthatanglescanbecopied.Let A and B bepointson l and n respectivelyandlet m bethelinethrough A and B .If t = m wearedone.Otherwise,let D beapointon t that isonthesamesideof n as l .(Assumingthestandardpropertiesof betweenness)Then, ∠BAD issmallerthantheangleat A formedby m and n.ByTheorem2.9weknowthattheinterioranglesat B and A sumtotworightangles,so ∠CBA and ∠BAD sumtolessthantwo rightangles.ByEuclid’sfifthpostulate t and l mustmeet.
2.1.6 Giventheassumptionsstatedintheexercise,ifwecopy ∠CBA to A,creatingline n,thenbyTheorem2.8 n and l willbe parallel.Also,thesumoftheinterioranglesat B and A forlines l and n willsumtotworightangles.Thus,lines t and n cannotbecoincident. Thus,byPlayfairline t cannotbeparallelto l Toshowthat t and l intersectonthesamesideof m as D and C ,weassumethattheyintersectontheotherside,atsomepoint E . Let F beapointon n thatisontheothersideof m from C ,andlet G apointon t onthissameside.Then, ∠BAF islessthan ∠BAG in measure,andsince ∠BAF ∼ = ∠CBA byTheorem2.9,wehavethatthe exteriorangle ∠CBA to∆BAE issmallerthananoppositeinterior angle(∠BAG),whichcontradictstheExteriorAngleTheorem.
Instructor’sGuidefor ExploringGeometry,SecondEdition
2.1.7 First,assumePlayfair’sPostulate,andletlines l and m be parallel,withline t perpendicularto l atpoint A.If t doesnotintersect m then, t and l arebothparallelto m,whichcontradictsPlayfair. Thus, t intersects m andbyTheorem2.9 t isperpendicularatthis intersection.
Now,assumethatwheneveralineisperpendiculartooneoftwo parallellines,itmustbeperpendiculartotheother.Let l bealineand P apointnoton l .Supposethat m and n arebothparallelto l at P Let t beaperpendicularfrom P to l .Then, t isperpendicularto m and n at P .ByTheorem2.4itmustbethat m and n arecoincident.
2.1.8 Tocreateaparallelto ←→ BC at A wecouldjustcopy ∠CBA to A anduseTheorem2.8.Thethreeanglesdefinedbythetriangleat A sumtotworightangles,andbyTheorem2.9wehavethatthesum oftheseanglesequalsthesumoftheanglesinthetriangle.
2.1.9 AssumePlayfairandletlines m and n beparalleltoline l .If m = n and m and n intersectat P ,thenwewouldhavetwodifferent linesparallelto l through P ,contradictingPlayfair.Thus,either m and n areparallel,orarethesameline.
Conversely,assumethattwolinesparalleltothesamelineareequal orthemselvesparallel.Let l bealineandsuppose m and n areparallel to l atapoint P noton l .Then, n and m mustbeequal,asthey intersectat P
2.1.10 AssumePlayfairandletline t intersectoneoftheparallel lines m and n,sayitintersects m at P .If m didnotintersect n,then t and m wouldbetwodifferentlinesbothparallelto n at P ,which contradictsPlayfair.
Conversely,assumethatifalineintersectsoneoftwoparallellines, itmustintersecttheother.Let l bealineandsuppose m and n are parallelto l atapoint P noton l ,with m = n.Then, m intersects n,whichisparallelto l .Byassumption, m mustintersect l ,andthus cannotbeparalleltol.
2.2CongruentTrianglesandPasch’sAxiom
ThissectionintroducesmanyresultsconcerningtrianglesandalsodiscussesseveralaxiomaticissuesthatarosefromEuclid’streatmentof triangles.
ThismaybeagoodpointtoreviewEuclid’s“proof”ofSAScongruence.Aninterestingdiscussionpointwouldbetohavestudentsvoice theiropinionastowhethertheproofwasvalidornot.
EuclideanGeometry 15
Also,thehistoryofaxiomsystemswouldbeagoodsupplemental activityatthispoint.Hilbert’saxiomsdidnotariseovernight.Hetook thebestofthosewhocamebeforehim,includingPasch,andmolded theseseparatestrandsintoacompletesystem.
2.2.1 Yes,itcouldpassthroughpoints A and B of∆ABC .Itdoes notcontradictPasch’saxiom,astheaxiomstipulatesthattheline cannotpassthrough A, B ,or C
2.2.2 Constructthediagonal AC oftherectangle ABCD .Then, alinepassingthroughasideoftherectanglewillbealinepassing throughasideofoneofthetwotrianglesdefinedbythediagonal andtheoriginalsidesoftherectangle.ByPasch’saxiom,thislinewill eitherpassthroughoneoftheothersidesofthetriangle,whichinclude therectanglesidesandthediagonal.Ifitpassesthroughasideofthe rectangle,wearedone.Ifitpassesthroughthediagonal,thenusing Pasch’saxiomasecondtime,wegetthatitmustpassthroughoneof theothertwosidesoftheothertriangle,andthusthroughasideof therectangle.
Thesameargumentcanbeusedrepeatedlytoshowthataline passingthroughaside(butnotavertex)ofanarbitraryn-gon(and notjustaregularn-gon)willintersectaside.Justpickavertexand constructinteriortrianglesbytakingalldiagonalsfromthisvertex.
2.2.3 No.Hereisacounter-example.
2.2.4 If A = B ,or B = C ,or A = C theresultfollowsimmediately. Otherwise,wecanassumeallthreepointsaredistinct. Iftheyalllieonaline,then,oneisbetweentheothertwo.Inevery case,wegetthat AC cannotintersect l .
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Instructor’sGuidefor ExploringGeometry,SecondEdition
Assumethepointsarenon-collinear.andthat A and C areon oppositesidesof l .Then, l intersects AC anddoesnotpassthrough A or C .ByPasch’saxiom,itmustintersect AB or BC atapointother than A, B ,or C .Sayitintersects AB .Thiscontradictstheassumption that A and B areonthesameside.Likewise,ifitintersects BC weget acontradiction.Thus, A and C areonthesamesideof l
2.2.5 If A = C wearedone.If A, B ,and C arecollinear,then B cannotbebetween A and C ,forthenwewouldhavetwopoints ofintersectionfortwolines.If A isbetween B and C ,then l cannot intersect AC .Likewise, C cannotbebetween A and B
Ifthepointsarenotcollinear,suppose A and C areonopposite sides.Then l wouldintersectallthreesidesof∆ABC ,contradicting Pasch’saxiom.
2.2.6 Apoint P isintheinteriorofa∆ABC if P isintheinterior of ∠ABC andintheinteriorof ∠BCA andintheinteriorof ∠CAB
2.2.7 Let ∠ABC ∼ = ∠ACB in∆ABC .Let −−→ AD betheanglebisector of ∠BAC meetingside BC at D .Then,byAAS,∆DBA and∆DCA arecongruentand AB ∼ = AC
2.2.8 ReferringtoFig.2.1,wecanusetheSSStrianglecongruence theoremon∆ADE and∆ABE toshowthat ∠EAB ∼ = ∠BAE
2.2.9 Supposethattwosidesofatrianglearenotcongruent.Then, theanglesoppositethosesidescannotbecongruent,asiftheywere, thenbythepreviousexercise,thetrianglewouldbeisosceles.
Supposein∆ABC that AC isgreaterthan AB .On AC wecan findapoint D between A and C suchthat AD ∼ = AB .Then, ∠ADB isanexteriorangleto∆BDC andisthusgreaterthan ∠DCB .But, ∆ABD isisoscelesandso ∠ADB ∼ = ∠ABD ,and ∠ABD isgreater than ∠DCB = ∠ACB
2.2.10 InthefigureaccompanyingTheorem2.11,supposethat BC wasgreaterthan YZ .Then,wecouldfindapoint D between B and C suchthat BD ∼ = YZ ,andbySAS∆ABD wouldbecongruentto ∆XYZ .Thisimpliesthat ∠BAD ∼ = ∠YXZ .But,wearegiventhat ∠BAC ∼ = ∠YXZ ,andso ∠BAD ∼ = ∠BAC .Thisimpliesthat D lies on ←→ AC ,andthat A, B , C arecollinear,whichisimpossible.
2.2.11 Let∆ABC and∆XYZ betworighttriangleswithright anglesat A and X ,andsuppose BC ∼ = YZ and AC ∼ = XZ .Suppose AB isgreaterthan XY .Then,wecanfindapoint D between A and B suchthat AD ∼ = XY .BySAS∆ADC ∼ = ∆XYZ .Now, ∠BDC isexteriorto∆ADC andthusmustbegreaterthan90degrees.But, ∆CDB isisosceles,andthus ∠DBC mustalsobegreaterthan90
EuclideanGeometry 17 degrees.Thisisimpossible,asthen∆CDB wouldhaveanglesum greaterthan180degrees.
FIGURE2.3:
2.2.12 Showthat∆ADB and∆BCA arecongruent,andthenshow that∆ADC and∆BDC arecongruent.
2.2.13 WeuseAAStoshowthat∆BFH ∼ = ∆AFG and∆CEI ∼ = ∆AEG.Thus BH ∼ = AG ∼ = CI and BHIC isSaccheri.Also,byadding congruentanglesintheleftcasewegetthatthesumoftheanglesin thetriangleisthesameasthesumofthesummitangles.Intheright case,weneedtore-arrangecongruentangles.
2.2.14 AssumePlayfairandlet ABCD beaSaccheriQuadrilateral withbase AB .ByTheorem2.8weknowthat ←→ AD isparallelto ←→ BC .By Theorem2.9thesummitanglesmustaddto180degrees.This,each anglemustbe90degrees.
Conversely,assumethatthesummitanglesofaSaccheriquadrilateralarealwaysrightangles.Let∆ABC beatriangle.Bytheprevious exercise,weknowthatwecanconstructaSaccheriquadrilateralbased onthetrianglewhosesummitanglesaddtotheanglesumofthetriangle.Thus,thesumoftheanglesinatriangleisalways180degrees. ByExercise2.1.8,thisimpliesPlayfair’saxiomistrue.
2.2.15 Givenquadrilaterals ABCD and WXYZ wesaythetwo quadrilateralsarecongruentifthereissomewaytomatchverticesso thatcorrespondingsidesarecongruentandcorrespondinganglesare congruent.
SASASTheorem:If AB ∼ = WX , ∠ABC ∼ = ∠WXY , BC ∼ = XY , ∠BCD ∼ = ∠XYZ ,and CD ∼ = YZ ,thenquadrilateral ABCD iscongruenttoquadrilateral WXYZ .
Instructor’sGuidefor ExploringGeometry,SecondEdition
Proof:∆ABC and∆WXY arecongruentbySAS.Thisimpliesthat∆ACD and∆WYZ arecongruent.Thisshowsthatsides arecorrespondinglycongruent,andtwosetsofanglesarecongruent (∠ABC ∼ = ∠WXY and ∠CDA ∼ = ∠YZW ).Since ∠BAC ∼ = ∠XWY and ∠CAD ∼ = ∠YWZ ,thenbyangleaddition ∠BAD ∼ = ∠XWZ Similarly, ∠BCD ∼ = ∠XYZ
2.3Project3-SpecialPointsofaTriangle
Studentsshouldbeencouragedtoexploreandexperimentinthislab project.Askthemifthereareanyothersetsofintersectinglinesthat onecouldconstruct.Or,arethereinterestingpropertiesofconstructed intersectinglinesinotherpolygons?
Someofthestudentsintheclasswillbefuturesecondarymath teachers.Thisprojectisonethatcouldbeeasilytransferredtothe highschoolsetting.Studentscouldhaveasanextracreditexercisethe taskofpreparingasimilarprojectforahighschoolclass.
2.3.1 ∆DGB and∆DGA arecongruentbySAS,asare∆EGB and∆EGC .Thus, AG ∼ = BG ∼ = CG.BySSS∆AFG ∼ = ∆CFG and sincetheanglesat F mustaddto180degrees,theanglesat F must becongruentrightangles.
2.3.2 Weprovedinthepreviousexercisethat,if G isthecircumcenter,then AG ∼ = BG ∼ = CG.Thus,thecirclecenteredat G with radius AG mustpassthroughtheothertwovertices.
2.3.3 Theanglepairsinquestionareallpairsofanexteriorangle andaninteriorangleonthesamesideforalinefallingontwoparallel lines.ThesearecongruentbyTheorem2.9.
Since ∠DAB , ∠BAC ,and ∠CAE sumto180degrees,and ∠BDA, ∠BAD ,and ∠ABD sumto180then,usingthecongruencesshown
EuclideanGeometry 19
inthediagram,wegetthat ∠DBA ∼ = ∠BAC .Likewise, ∠BAD ∼ = ∠ABC .ByASAwegetthat∆ABC ∼ = ∆BAD .Similarly,∆ABC ∼ = ∆CEA and∆ABC ∼ = ∆FCB
2.3.4 Analtitudeof∆ABC willmeetasideatrightangles.Thus, thealtitudemeetsthesideoftheassociatedtriangleatrightangles, asthissideisparalleltothesideof∆ABC .Also,bythetriangle congruenceresultofthepreviousexercise,thealtitudebisectstheside oftheassociatedtriangle.
2.3.5 Let −−→ AB and −→ AC defineanangleandlet −−→ AD bethebisector. Dropperpendicularsfrom D to −−→ AB and −→ AC ,andassumetheseintersect at B and C .Then,byAAS,∆ABD and∆ACD arecongruent,and BD ∼ = CD
Conversely,suppose D isinteriorto ∠BAC with BD perpendicular to −−→ AB and CD perpendicularto −→ AC .Also,supposethat BD ∼ = CD Then,bythePythagoreanTheorem AB 2 + BD 2 = AD 2 and AC 2 + CD 2 = AD 2 .Thus, AB ∼ = AC andbySSS∆ABD ∼ = ∆ACD .This impliesthat ∠BAD ∼ = ∠CAD .
2.4.1Mini-Project:AreainEuclideanGeometry
Thissectionincludesthefirst“mini-project”forthecourse.These projectsaredesignedtobedoneintheclassroom,ingroupsofthree orfourstudents.EachgroupshouldelectaRecorder.TheRecorder’s solejobistooutlinethegroup’ssolutionstoexercises.Thesummary shouldnotbeaformalwrite-upoftheproject,butshouldgiveenough abriefsynopsisofthegroup’sreasoningprocess.
Themaingoalforthemini-projectsistohavestudentsdiscuss geometricideaswithoneanother.Throughthegroupprocess,students clarifytheirownunderstandingofconcepts,andhelpeachotherbetter graspabstractwaysofthinking.Thereisnobetterwaytoconceptualize anideathantohavetoexplainittoanotherperson.
Inthismini-project,studentsareaskedtograpplewiththenotion of“area”.Youmaywanttoprecedetheprojectbyageneraldiscussion ofhowtobestdefinearea.Studentswillquicklyfindthatitisnotsuch anobviousnotionastheyoncethought.Forexample,whatdoesit meanfortwofigurestohavethesamearea?
2.4.1 Constructadiagonalandusethefactthatalternateinterior anglesofalinefallingonparallellinesarecongruenttogeneratean ASAcongruenceforthetwosub-trianglescreatedintheparallelogram.
2.4.2 Weknowthat AD ∼ = EF .Thus, AE ∼ = DF ,asthelengthof
20 Instructor’sGuidefor ExploringGeometry,SecondEdition
eitherofthesedifferbythelengthof DE .Also, ∠AEB ∼ = ∠DFC ,by Theorem2.9.BySASwegetthat∆AEB ∼ = ∆DFC .Thus,parallelogram ABCD canbesplitinto∆AEB and EBCD andparallelogram EBCF canbesplitinto∆DFC and EBCD .Clearly,thesearetwo pairsofcongruentpolygons.
2.4.3 Wehavethat AE ∼ = DF .Theorem2.9saysthat ∠AEB ∼ = ∠DFC .So,bySAS∆AEB ∼ = ∆DFC .Let G bethepointwhere CD intersects BE .(SuchapointexistsbyPasch’saxiomappliedto ∆AEB )Now,parallelogram ABCD canbesplitinto∆AEB plus ∆BGC minus∆DGE .Also,parallelogram EBCF canbesplitinto ∆DFC plus∆BGC minus∆DGE
2.4.4 UseTheorem2.8andExercise2.4.1.
2.4.5 ByTheorem2.9weknowthat ∠BAE and ∠FBA areright angles,andthus ABFE isarectangle.ByTheorem2.9wehave that ∠DAB ∼ = ∠CBG,where G isapointon −−→ AB totherightof B .Subtractingtherightangles,weget ∠DAE ∼ = ∠CBF .BySAS, ∆DAE ∼ = ∆CBF .Thenrectangle AEFB canbesplitinto AECB and∆CBF andparallelogram DABC canbesplitinto AECB and ∆DAE andthefiguresareequivalent.
HiddenAssumptions?Onehiddenassumptionisthenotionthat areasareadditive.Thatis,ifwehavetwofiguresthatarenotoverlapping,thentheareaoftheunionisthesumoftheseparateareas.
2.4.2CeviansandArea
2.4.6 Sinceamedianisaceviantoamidpoint,thenthefractionsin theratioproductofTheorem2.24areallequalto1.
2.4.7 LetthetriangleandmediansbelabeledasinTheorem2.24. Theareaof∆AYB willbeequalto AYh,where h isthelengthof aperpendiculardroppedfrom B to ←→ AC .Theareaof∆CYB willbe equalto CYh,Since AY ∼ = CY ,theseareaswillbethesameand ∆ABC willbalancealong ←→ BY .Asimilarargumentshowsthat∆ABC balancesalongeachmedian,andthusthecentroidisabalancepoint forthetriangle.
2.4.8 Refertothefigurebelow.Bythepreviousexerciseweknow that1+2+3=4+5+6(intermsofareas).Also,since1and2share thesamebaseandheightwehave3=4.Similarly,1=2and5=6. Thus,1=6.
Similarly,2+3+4=1+5+6willyield4=5,and3+4+5=1+5+5 yields2=3.Thus,all6havethesamearea.
FIGURE2.5:
2.4.9 Considermedian BD in∆ABC ,with E thecentroid.Let a = BE and b = DE .Then,theareaof∆DBC is (a+b)h 2 where h is theheightofthetriangle.Thisis3timestheareaof∆DEC bythe previousexercise.Thus, (a
,or
2.5SimilarTriangles
Asstatedinthetext,similarityisoneofthemostusefultoolsinthe geometer’stoolkit.Severaloftheexercisescouldbejumpingoffpoints forfurtherdiscussion—thedefinitionofthetrigonometricfunctions, thePythagoreanTheorem(thiscouldbeaplacetoreviewsomeof themyriadofwaysthatthistheoremhasbeenproved),andPascal’s TheoremanditsuseinHilbert’sdevelopmentofarithmetic.
2.5.1 Since ←→ DE cutstwosidesoftriangleatthemidpoints,then byTheorem2.27,thislinemustbeparalleltothethirdside BC .Thus ∠ADE ∼ = ∠ABC and ∠AED ∼ = ∠ACB .Sincetheangleat A iscongruenttoitself,wehavebyAAAthat∆ABC and∆ADE aresimilar, withproportionalityconstantof 1 2 .
Instructor’sGuidefor ExploringGeometry,SecondEdition
FIGURE2.7:
2.5.2 Since AG ∼ = DE then AB AG = AB DE andso AB AG = AC DF .Since AH ∼ = DF weget AB AG = AC AH .ByTheorem2.27, ←→ GH and ←→ BC are parallel.
2.5.3 Let∆ABC and∆DEF havethedesiredSSSsimilarityproperty.Thatissides AB and DE ,sides AC and DF ,andsides BC and EF areproportional.Wecanassumethat AB isatleastaslargeas DE .Let G beapointon AB suchthat AG ∼ = DE .Let ←→ GH bethe parallelto ←→ BC through G.Then, ←→ GH mustintersect ←→ AC ,asotherwise ←→ AC and ←→ BC wouldbeparallel.Bythepropertiesofparallels, ∠AGH ∼ = ∠ABC and ∠AHG ∼ = ∠ACB .Thus,∆AGH and∆ABC aresimilar.
Therefore, AB AG = AC AH .Equivalently, AB DE = AC AH .Wearegiventhat AB DE = AC DF .Thus, AH ∼ = DF
Also, AB AG = BC GH and AB AG = AB DE = BC EF .Thus, GH ∼ = EF BySSS∆AGH and∆DEF arecongruent,andthus∆ABC and ∆DEF aresimilar.
FIGURE2.8:
2.5.4 Since ∠ACD and ∠DCB havemeasuressummingto90
EuclideanGeometry 23
degrees,andsince ∠DAC and ∠ACD sumto90,then ∠DCB ∼ = ∠CAD (∼ = ∠CAB ).Likewise, ∠CBD (∼ = ∠ABC ) ∼ = ∠ACD .ByAAA ∆DCB and∆CAB aresimilar,asare∆ACD and∆ABC .Thus, y a = a c and x b = b c .Or, y = a 2 c and x = b2 c .Thus, c = x + y = a 2 +b2 c . Theresultfollowsimmediately.
2.5.5 Anyrighttriangleconstructedsothatoneangleiscongruent to ∠A musthavecongruentthirdangles,andthustheconstructed trianglemustbesimilarto∆ABC .Since sin and cos aredefinedin termsofratiosofsides,thenproportionalsideswillhavethesameratio, andthusitdoesnotmatterwhattriangleoneusesforthedefinition.
2.5.6 Dropaperpendicularfrom C to ←→ AB intersectingat D .There aretwocases.If D isnotbetween A and B ,thenitistototheleft of A ortotherightof B .Wecanassumeitistotheleftof B .Then, theangleat A mustbeobtuse,as ∠BAC isexteriortorighttriangle ∆ACD .If D isbetween A and B thenwecanassumetheanglesat A and B areacute,againbyanexteriorangleargument. Inthefirstcase, sin(∠A)= CD b and sin(∠B )= CD a .Then, sin(∠A) sin(∠B ) = a b Inthesecondcase, sin(∠A)= sin(∠DAC ).Anexactlyanalogous argumenttothefirstcasefinishestheproof.
FIGURE2.9:
2.5.7 Iftheparallelto ←→ AC doesnotintersect ←→ RP ,thenitwould beparalleltothisline,andsinceitisalreadyparallelto ←→ AC ,thenby Exercise2.1.9 ←→ RP and ←→ AC wouldbeparallel,whichisimpossible. Bythepropertiesofparallels, ∠RAP ∼ = ∠RBS and ∠RPA ∼ = ∠RSB .Thus,byAAA∆RBS and∆RAP aresimilar.∆PCQ and
24 Instructor’sGuidefor ExploringGeometry,SecondEdition
∆SBQ aresimilarbyAAAusingananalogousargumentfortwoof theanglesandtheverticalanglesat Q
Thus, CP BS = CQ BQ = PQ QS ,and AP BS = AR BR = PR SR .So, CP AP BQ QC = CP AP BS CP = BS AP And, CP AP BQ QC AR RB = BS AP AR RB = BS AP AP BS =1.
2.5.8 ByTheorem2.25appliedto∆ADE and∆ABC weget AD AB = AE AC .Again,usingTheorem2.25on∆AFE and∆ABG weget AF AB = AE AG .Thus, AD AF = AG AC
2.5.1Mini-Project:FindingHeights
Thismini-projectisagoodexampleofanactivityfuturehighschool geometryteacherscouldincorporateintotheircourses.Itisavery practicalapplicationofthenotionofsimilarity.Themathematicsinthe firstexampleforfindingheightisextremelyeasy,buttheinteresting partisthedatacollection.Studentsneedtodeterminehowtogetthe mostaccuratemeasurementsusingthematerialstheyhaveonhand. Thesecondmethodoffindingheightisagainasimplecalculation usingtwosimilartriangles,butthestudentsmaynotseethisatfirst. Theinterestingpartoftheprojectishavingthemseetheconnection betweenthemirrorreflectionandthecalculationtheymadeinpartI. Again,havethestudentsworkinsmallgroupswithaRecorder,but makesuretheRecorderpositiongetsshiftedaroundfromprojectto project.
2.6CircleGeometry
Thissectionintroducesstudentstothebasicgeometryofthecircle. Thepropertiesofinscribedanglesandtangentsarethemostimportant propertiestofocusoninthissection.
2.6.1 Case2: A isonthediameterthrough OP .Let α = m∠PBO and β = m∠POB .Then, β =180 2α.Also, m∠AOB =180 β =2α. Case3: A and B areonthesamesideof ←→ PO .Wecanassumethat m∠OPB>m∠OPA.Let m∠OPB = α and m∠OPA = β .Then,we canargueinasimilarfashiontotheproofoftheTheoremusing α β insteadof α + β
2.6.2 Letquadrilateral ABCD beinscribedinthecircle,withcenter O .Then,byTheorem2.31 a = ∠OAB = 1 2 ∠EOB ,where E is thepointofintersectionof −→ AO withthecircle.Also, b = ∠OAD = 1 2 ∠EOD
FIGURE2.10:
Likewise,wewouldhavethisrelationshipforanglesgeneratedby−−→ CO .
FIGURE2.11:
Clearly,thesumoftheanglesat O (2a +2b +2c +2d)is360,and sothesumoftheanglesat A and C is180.
2.6.3 Consider ∠AQP .ThismustbearightanglebyCorollary2.33.Similarly, ∠BQP mustbearightangle.Thus, A, Q,and B arecollinear.
2.6.4 ByTheorem2.31weknowthat m∠AOP =2m∠ABP and m∠POC =2m∠PBC .But, −−→ BP bisects ∠ABC andso ∠AOP ∼ = ∠POC .Let Q bethepointofintersectionof OP and AC .Then, ∆OQA ∼ = ∆OQC bySAS.Theresultfollows.
2.6.5 Let AB bethechord, O thecenter,and M themidpointof AB .Then∆AOM ∼ = ∆BOM bySSSandtheresultfollows.
Instructor’sGuidefor ExploringGeometry,SecondEdition
2.6.6 ∠BAD ∼ = ∠BCD byCorollary2.32.Likewise, ∠CBA ∼ = ∠CDA.Thus,∆ABP and∆CDP aresimilar.Theresultfollowsimmediately.
2.6.7 Consideratriangleonthediagonaloftherectangle.Thishas arightangle,andthuswecanconstructthecircleonthisangle.Since theothertriangleintherectanglealsohasarightangleonthesame side(thediameterofthecircle)thenitisalsoinscribedinthesame circle.
2.6.8 First, m∠BDA + m∠CAD =180 m∠DPA = m∠CPD . Then,byTheorem2.31,wehave m∠BDA + m∠CAD = 1 2 (m∠BOA + m∠COD ).Since m∠CPD = m∠BPA (Verticalangles),theresult follows.
2.6.9 Ifpoint P isinsidethecircle c,thenTheorem2.41applies. But,thistheoremsaysthat m∠BPA = 1 2 (m∠BOA + m∠COD ),where C and D aretheotherpointsofintersectionsof ←→ PA and ←→ PB with thecircle.If P isinside c,then C and D aredifferentpoints.The assumptionofTheorem2.42saysthat m∠BPA = 1 2 m∠BOA.But, m∠BPA = 1 2 (m∠BOA + m∠COD )wouldthenimplythat m∠COD = 0,whichisimpossibleas C and D arenotcollinearwith O
2.6.10 ByTheorem2.36 m∠PTA =90 m∠ATO .Sincetriangle OATisisosceles(OA and OT areradiiof c),then ∠ATO ∼ = ∠OAT Since m∠TOA =180 (m∠ATO + m∠OAT =180 2m∠ATO ,then m∠ATO = 1 2 (180 m∠TOA =90 1 2 m∠TOA).
So, m∠PTA =90 m∠ATO =90 (90 1 2 m∠TOA)= 1 2 m∠TOA.
2.6.11 Theanglemadeby BT and l mustbearightangleby Theorem2.36.Likewise,theanglemadeby AT and l isarightangle. Thus, A, T ,and B arecollinear.
2.6.12 Supposetheyintersectedatanotherpoint P .Then,∆TBP and∆TAP arebothisoscelestriangles.But,thiswouldimply,bythe previousexercise,thatthereisatrianglewithtwoanglesgreaterthan arightangle,whichisimpossible.
2.6.13 Supposeoneofthecircleshadpoints A and B onopposite sidesofthetangentline l .Then AB wouldintersect l atsomepoint P whichisinteriortothecircle.But,then l wouldpassthroughan interiorpointofthecircleandbycontinuitymustintersectthecircle intwopointswhichisimpossible.Thus,eitherallpointsofonecircle areonoppositesidesof l fromtheothercircleorareonthesameside.
2.6.14 Let P and Q bepointsonthetangent,asshown.Then, ∠BDT ∼ = ∠BTP ,asbothareinscribedanglesonthesamearc.Like-
EuclideanGeometry 27
wise, ∠ACT ∼ = ∠ATQ.Since, ∠BTP ∼ = ∠ATQ (verticalangles),then ∠BDT ∼ = ∠ACT andthelines ←→ AC and ←→ BD areparallel.
2.6.15 ByTheorem2.36,wehavethat ∠OAP isarightangle,as is ∠OBP .Sincethehypotenuse(OP )andleg(OA)ofrighttriangle ∆OAP arecongruenttothehypotenuse(OP )andleg(OB )ofright triangle∆OBP ,thenbyExercise2.2.10thetwotrianglesarecongruent.Thus ∠OPA ∼ = ∠OPB .
2.6.16 Supposethatthebisectordidnotpassthroughthecenter. Then,constructasegmentfromthecentertotheoutsidepoint.Bythe previoustheorem,thelinecontinuedfromthissegmentmustbisectthe anglemadebythetangents.But,thebisectorisunique,andthusthe originalbisectormustpassthroughthecenter.
2.6.17 Let A and B bethecentersofthetwocircles.Construct thetwoperpendicularsat A and B to ←→ AB andlet C and D bethe intersectionswiththecirclesononesideof ←→ AB . If ←→ CD doesnotintersect ←→ AB ,thentheselinesareparallel,andthe anglesmadeby ←→ CD andtheradiiofthecircleswillberightangles. Thus,thislinewillbeacommontangent.
Otherwise, ←→ CD intersects ←→ AB atsomepoint P .Let ←→ PE beatangent tothecirclewithcenter A.Then,since∆PAC and∆PBD aresimilar, wehave AP BP = AC BD .Let ←→ BF beparallelto ←→ AE with F theintersection oftheparallelwiththecirclecenteredat B .Then, AC BD = AE BF .So, AP BP = AE BF .BySASsimilarity,∆PAE and∆PBF aresimilar,andso F ison ←→ PE and ∠PFB isarightangle.Thus, ←→ PE isatangenttothe circlecenteredat B .
28 Instructor’sGuidefor ExploringGeometry,SecondEdition
2.7Project4-CircleInversionandOrthogonality
ThissectioniscrucialforthelaterdevelopmentofthePoincar´emodel ofnon-Euclidean(hyperbolic)geometry.Itisalsohassomeofthemost elegantmathematicalresultsfoundinthecourse.
2.7.1 ByTheorem2.31, ∠Q2 P1 P2 ∼ = ∠Q2 Q1 P2 .Thus, ∠PP1 Q2 ∼ = ∠PQ1 P2 .Sincetriangles∆PP1 Q2 and∆PQ1 P2 sharetheangleat P ,thentheyaresimilar.Thus, PP1 PQ1 = PQ2 PP2 ,or(PP1 )(PP2 )= (PQ1 )(PQ2 ).
2.7.2 Choosealinefrom P passingthroughthecenter.Then, PP1 PP2 =(PO OP1 )(PO + OP1 )= PO 2 r 2 .
2.7.3 Bysimilartriangles OP OT = OT OP .Since OT = r theresult follows.
2.7.4 As P approaches O ,thedistance OP goestozero,sothe distance OP mustgetlargerwithoutbound,fortheproducttoremain equalto r 2 .Thus,theorthogonalcircleradiusgrowslargerwithout boundaswell.