Transformational Proof in High School Geometry · 2018-03-30 · Preface p. 1 Preface:...

Transformational Proof in High School Geometry

A guide for teachers and curriculum developers

Lew Douglas and Henri Picciotto

March 2018

The latest version of this document will always be at http://www.MathEducationPage.org/transformations/

Preface: Rationale for Transformational Proof

Chapter 1: Isometries and Congruence

Chapter 2: Symmetry Definitions and Properties: Triangles and Quadrilaterals

Chapter 3: Proving Triangles and Quadrilaterals Satisfy Symmetry Definitions

Chapter 4: Dilation and Similarity

Chapter 5: Circles

Chapter 6: The Pythagorean Theorem

Appendix: List of Theorems in Chapters 1, 4, and 5

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March2018 ©LewDouglasandHenriPicciotto

www.MathEducationPage.org Prefacep.1

Preface:TransformationalProofRationale

TheCommonCoreStateStandardsforMathematics(CCSSM)includeafundamentalchange

inthegeometrycurriculumingrades8through10:geometrictransformations,not

congruenceandsimilaritypostulates,aretoconstitutethelogicalfoundationofgeometry

atthislevel.Weproposeanapproachtotrianglecongruenceandsimilarity,andmore

generallytogeometricproofwhereadvantageous,thatiscompatiblewiththisnewvision.

Apedagogicalargumentforthischange:Congruencepostulatesarerathertechnicalandfar

fromself-evidenttoabeginner.Infact,manyteachersintroducethebasicideaof

congruencebysayingsomethinglike“Ifyoucansuperposetwofigures,theyare

congruent.”Thatisnotveryfarfromsaying“Ifyoucanmoveonefiguretolandexactlyon

topoftheother,theyarecongruent.”Inotherwords,basingcongruenceon

transformationsismoreintuitivethangoingintheotherdirection.

Therearealsomathematicalargumentsforthechange.Atransformationalapproach

1. offersdeeperlinksbetweenalgebraandgeometrybecauseofitsemphasisonfunctions(andthuscompositionoffunctions,inversefunctions,fixedpoints,andsoon.)

2. highlightsthenaturalwaytransformationsconnectwithcomplexnumbersandmatrices,socanimmeasurablyenhancetheteachingofthesetopicsingrades11-12.

3. willgivesymmetryagreaterroleinschoolmathematics,whichnotonlyenhancesgeometricthinkingandconnectseasilywithartandnature,butalsoenhancesstudent

motivation.

4. makesitpossibletodiscussthesimilarityofcurves(suchascirclesandparabolas),whichcouldnotbedoneunderthetraditionaldefinitionofsimilaritybecauseitrelies

onequalanglesandproportionalsides.

5. pavesthewayfortransformationsofgraphsandtransformationsusingcoordinatesinintermediatealgebraandbeyond.

6. blendsmorenaturallywithdynamicgeometrysoftware,suchasthegeometrycomponentofDesmosandGeoGebra.Thisisbecausethetransformationstoolsare

oftenusefulinconstructingdynamicspecialpolygons.

Oneoftheconsequencesofthischangeistheneedforsomeclarityonhowthisnewvision

affectsthelogicalstructureofhighschoolgeometry.Ourdevelopmentsofarattemptstofill

thatneedbystartingwithanewsetofdefinitionsandassumptions,thenusingtheseto

provetheusual(andsomelessusual)theorems.Whenwecompletethisdevelopment,

meantsolelyforteachersandcurriculumdevelopers,wehopetobeginworkonmaterials

forstudents.Ifyouusethisdocumentasaguideforworkwithstudents,orifyouhave

commentsorquestions,pleaseletusknow:[email protected]

[email protected].

Beyondpedagogicalandmathematicalconsiderations,transformationsareanintegralpart

ofeverydaylife.Flipsandturnsarebuilt-intoolswhendrawingshapesonacomputer.

Enlargementorreductiononacomputerscreenoracopierisadilationwiththescale

factorshownasapercent.Whenweseeaphotograph,weknowthatthepeopledepictedin


www.MathEducationPage.org Prefacep.2

itarenotreally2incheshigh–theirimageshavebeendilatedusingapositivescalefactor

lessthan1.Imagesonamoviescreenhavebeendilatedusingascalefactorgreaterthan1.

Flatmapsaretransformationsofthesurfaceofaspheretoaplane.Aswelookaroundus,

moreandmoreexamplesoftransformationsbecomeapparent.

NoteonGeometricConstruction

Theessentialmathematicalconceptunderlyinggeometricconstructionisnottheuseofstraightedgeandcompass.Interestingversionsofconstructionhavebeendevelopedfor

straightedgeandthecollapsingcompass,andforthecompassalone,nottomentionfor

pedagogicaltoolssuchaspattypaper,Plexiglasmirrors,andofcourseinteractivegeometry

software.

Theessentialconceptunderlyinggeometricconstructionisthatofintersectingloci.Thelocusofapointisthesetofallpossiblelocationsofthatpoint,giventhepoint’sproperties.Thelocuscanbealine,acircle,orsomeothercurve.Ifoneknowstwolociforacertain

point,thepointmustlieattheirintersection.Inotherwords,givenafigure,anadditional

pointcanbeaddedtoitinamathematicallyrigorouswaybyknowingthelocus(location)

ofthepointintwodifferentways.Geometricconstructionisthechallengeoffindingsuch

pointsand,insomecases,usingthemtodefineadditionalpartsofthefigure.

Inthesequencewepropose,argumentsbasedonwhatwecalltheconstructionpostulates(seeChapter1)areneededformanyproofs.Onepedagogicalconsequenceofthisisthat

constructionchallengesshouldplayanimportantroleinteachinggeometry.Wefavoran

introductionusingcompass,straightedge,andpattypaper,soonfollowedbyworkwith

interactivegeometrysoftware.

NoteonCommonCorecompliance

TheCommonCoreStateStandardsforMathematicsonlyrequireusingtransformationsto

justifythetrianglecongruenceandsimilaritycriteria.Beyondthat,theydonotspecify

whetherproofscanorshouldusetransformationalapproaches.Themosttraditional

interpretationoftheStandardswouldreverttotraditionalproofsbasedoncongruentand

similartriangles.Theapproachwetakeinthisdocumentistouseatransformational

approachwheneverpossible.Classroomteachersandcurriculumdeveloperscansituate

themselvesanywhereinthisrange.Thebeststrategyisprobablyonethatcombines

traditionalandtransformationalapproachestoproofindifferentcases,andalsocompares

theminsomeinstances.Notethatwedidnotincludetheoremsforwhichwecouldnotfind

asuitabletransformationalalternativetothetraditionalproof.

MoreonTransformationalGeometry

Seehttp://www.mathedpage.org/transformationsforrelatedarticles,andsomecurricular

materials.


www.MathEducationPage.org Chapter1,p.1

Chapter1:IsometriesandCongruenceBasicTransformationalDefinitions

TransformationofthePlane:Aone-to-onefunctionwhosedomainandrangearetheentireplane.(Itisunderstoodthatweareworkinginthe2-Dplane.)Isometry(orrigidmotion):Atransformationoftheplanethatpreservesdistanceandangles.SymmetryofaFigure:Anisometryforwhichthefigure,takenasawhole,isinvariant.Individualpointsinthefigureneednotbefixed.Congruence:Twofiguresarecongruentifonecanbesuperposedontheotherbyasequenceofisometries.(Or:twofiguresarecongruentifoneistheimageoftheotherunderacompositionofisometries.)Note:Manytraditionaldefinitionsareunchanged.Forexample,theperpendicularbisectorofasegmentisthelineperpendiculartothesegmentthroughitsmidpoint.

NewDefinitionofParallel

Itwillsimplifythestatementofsometheoremsifwecalllinesthatcoincideparallel.Thisisespeciallyusefulfortheoremsinvolvingtranslation.Therefore,wewillsaythattwolinesareparalleliftheydonotintersectorcoincide.Wewillcalltraditionalparallellines“distinctparallellines.”Segmentsorraysareparallelifthelinesthatcontainthemareparallel.Parallelrayscanpointinthesameoroppositedirections.

NotationWeusethecustomarysymbolsforparallel,perpendicular,angle,andtriangle,butwegenerallydonotusesymbolsforsegment,ray,orline.Forexample,wesaysegmentABorlineCD.EFbyitselfcanbeanyoftheseifthecontextmakesitclear.Asiscustomary,EFcanalsobethedistancebetweenEandF,whichisalsothelengthofsegmentEF.Wesometimesuseanarrowoveroneormorelettersforavector,butcontextusuallymakesitclear.Ingeneral,werefertotheimageofpointPunderatransformationasP',andlikewiseforlinesandotherfigures.

DefinitionsofBasicIsometriesReflection:Areflectioninalinebmapsanypointonbtoitself,andanyotherpointPtoapointP'sothatbistheperpendicularbisectorofsegmentPP'.Rotation:GivenapointOandadirectedangleθ,theimageofapointP≠OunderarotationwithcenterOandangleθisapointP'onthecirclecenteredatOwithradiusOP,suchthat∠POP'=θ.TheimageofOisO.(θispositiveforacounterclockwiserotation,negativeforclockwise.)A180°rotationisalsoknownasahalf-turnorreflectioninapoint(thecenterofrotation).



VectorAnarrowspecifyingdistanceanddirectionforwhichpositiondoesn'tmatter.Tworepresentativesofthesamevectorwillbeparallel,havethesamelength,andpointinthesamedirection.! = !meanstheyaretworepresentativesofthesamevector.ParallelVectors:Vectorswhoserepresentativeslieonparallellines.Theydonotneedtohavethesamelength.OppositeVectors:Twoequal-lengthparallelvectorspointinginoppositedirections.−!denotesthevectoropposite!.Translation:Givenavector!,theimageofapointPunderatranslationby!,isapointP'suchthat!!′ = !.

Postulates

Thefollowingfiveassumptionsaresufficientforthemathematicallyexperienced,thoughatrulyrigorousdevelopmentwouldincludetheprotractorpostulateandothersfromHilbert'sset.Whenworkingwithstudentsordevelopingcurriculum,manyofthebasictheoremsprovedbelowcanbeaddedtothesetofassumptionsbecausemanystudentswillthinktheyareobvious.1. Theparallelpostulate:Throughapointoutsideagivenline,oneandonlyoneline

canbedrawnparalleltothegivenline.Note:Theparallelpostulatehasmanyequivalentforms.ThisoneisgenerallycreditedtoPlayfair.Seehttps://en.wikipedia.org/wiki/Parallel_postulate.

2. Reflectionpreservesdistanceandanglemeasure.Andtheconstructionpostulates:3. Twodistinctlinesmeetinatmostonepoint.4. Acircleandalinemeetinatmosttwopoints.5. Twodistinctcirclesmeetinatmosttwopoints.

PedagogicalNote

Inordertolimitourselvestoaminimumnumberofpostulates,weonlyassumethatreflectionisanisometry.Toensurethelogicalprogressionofthispresentation,weprovethatrotationsandtranslationsareisometriesfurtherdown(see“TwoReflections”below).However,intheclassroomitmaybepreferabletopresentallthreeasisometriesearlyon,andallowstudentstofindtransformationalproofsofTheorems9-12thatrelyononeormoreofthethreebasicisometries.Thiscanbedone,say,usinginteractivegeometrysoftware.Sometheoremsinthischapterareprobablyobvioustoyourstudents.Supplyingformalproofsforthemwilllikelybecounterproductive,asstudentsfinditincomprehensiblethatsuchresultsrequireproof.Thisistruewhethertheproofissimpleorcomplicated.Wehavefounditmoreeffectivetoreserveformalproofforlessobvioustheorems,suchasthoseinsubsequentchapters.Inthisdocument,weuseanasterisk(*)tomarktheoremsforwhichwediscourageaformaldiscussion.Forcompletenessandlogicalconsistency,weofferproofsforthesetheoremsbelow,butintheclassroom,youcanprobablyeitherassumethoseresultsordiscusstheminformally.Admittedly,thisisajudgmentcallineachcase.



Requiringaminimallistofaxiomsandprovingseeminglyobviousresultsdoesnotconstituteanaccessibleintroductiontoproof.Infact,itisarathersophisticatedstance.Shouldyouhaveanexceptionalstudentwhodemandsarigoroustreatmentwithaminimumnumberofassumptions,referthemtothisdocument.

BasicTheorems

1. *IfA'=Bunderareflection,thenB'=A.Proof:Iflisthereflectionline,thenlistheperpendicularbisectorofsegmentAB,whichisthesameassegmentBA.

2. *Reflectionpreservescollinearityandbetweenness.Proof:A,B,andCarecollinearwithBbetweenAandCifandonlyifAB+BC=AC.Sincereflectionpreservesdistance,A'B'+B'C'=A'C'.

3. *IfA→A'andB→B'underareflection,segmentABmustmapontosegmentA'B'.Proof:Becausereflectionpreservescollinearityandbetweenness,segmentABmustmapontopartorallofsegmentA'B'.ReflectanypointPonsegmentA'B'underthesamereflection.P'→PbyTheorem1.So,anypointonsegmentA'B'isanimagepointandsegmentABreflectsontotheentiresegmentA'B'.

4. *Reflectionsmapraysontotoraysandlinesontolines.Proof:TheargumentofTheorem3applies.

5. *Congruentsegmentshaveequallength.Congruentangleshaveequalmeasure.Proof:Isometriespreservesegmentlengthandanglemeasure.

6. *Thecorrespondingsidesandanglesofcongruentpolygonshaveequalmeasure.Proof:Isometriespreservesegmentlengthandanglemeasure.(ThisistrueforreflectionbyPostulate2.Itwillbeprovedforrotationandtranslationlaterinthisdocument.)

7. *ThereisareflectionthatmapsanygivenpointPontoanygivenpointQ.Proof:IfP=Q,reflectioninanylinethroughPwilldothejob.Ifnot,QisthereflectionofPacrosstheperpendicularbisectorofsegmentPQ.



TriangleCongruence8. ApointPisequidistantfromtwopointsAandBifandonlyifitliesontheir

perpendicularbisector.Proof:IfPisontheperpendicularbisectorbofAB,thenbythedefinitionofreflection,AandBareimagesofeachotherandPisitsownimageinareflectionacrossb.So,PA=PBsincereflectionspreservedistance.Conversely,ifPA=PB,wemustshowthatPisontheperpendicularbisectorofAB.LetMbethemidpointofAB.ReflectAacrossPM.CalltheimageA'.Sincereflectionspreservedistance,A'mustbeonthecirclecenteredatP,withradiusPA,andonthecirclecenteredatM,withradiusMA.BecausePA=PBandMisthemidpointofAB,Bmustbeonbothcirclesaswell.A'≠AbecauseAisnotonthereflectionline,soA'=B,theotherintersectionpointofthecircles.SincePMistheperpendicularbisectorofAA',itistheperpendicularbisectorofAB.WeconcludethatPisindeedontheperpendicularbisectorofAB.(Anindirectproofisalsopossibleifoneassumesthetriangleinequality.)

9. IftwosegmentsABandCDhaveequallength,thenoneistheimageoftheother,withCtheimageofAandDtheimageofB,undereitheroneortworeflections.Proof:GivenAB=CD,byTheorem7,wecanreflectsegmentABsothatCistheimageofA.LetB'betheimageofB.IfB’=D,thatsinglereflectionwilldo.Ifnot,sincereflectionspreservedistance,wehaveCB'=AB=CD.ByTheorem8,CisontheperpendicularbisectorbofB'D.Therefore,asecondreflectionofCB'inbyieldsCD.

10. Equallengthsegmentsarecongruent.IfwecombinethiswithTheorem5,wehave:Segmentsarecongruentifandonlyiftheyhaveequallength.Proof:ThisisanimmediatecorollaryofTheorem9.



11. CongruenceCriteriaforTrianglesa. SSSCongruence:Ifallsidesofonetrianglearecongruent,respectively,toall

sidesofanother,thenthetrianglesarecongruent.Proof:Wearegiven∆ABCand∆DEFwithAB=DE,BC=EF,andAC=DF.ByTheorem9,wecansuperposeABontoDEinoneortworeflections.Becausereflectionspreservedistance,C'(theimageofC)mustbeattheintersectionoftwocircles:onecenteredatD,withradiusDF,theothercenteredatE,withradiusEF.F,ofcourse,isonbothcircles.IfC'=F,we’redone.Ifnot,C'mustbeattheotherintersection.ButbyTheorem8,DEmustbetheperpendicularbisectorofFC',soareflectionacrossDEsuperposes∆ABConto∆DEF.

b. SASCongruence:Iftwosidesofonetrianglearecongruenttotwosidesofanother,andiftheincludedangleshaveequalmeasure,thenthetrianglesarecongruent.Proof:Wearegiven∆ABCand∆DEF,withAB=DE,BC=EF,and∠B=∠E.ByTheorem9,wecansuperposeABontoDEinoneortworeflections.IfC'=F,we’redone.Ifnot,reflectFacrossDEwithimageF'.Becausereflectionspreservedistanceandanglemeasure,C'mustlieontherayDF'andonthecirclecenteredatDwithradiusDF.ThereforeC'=F',soareflectionacrossDEsuperposes∆ABConto∆DEF.

c. ASACongruence:Iftwoanglesofonetrianglearecongruenttotwoanglesofanother,andifthesidescommontotheseanglesineachtrianglearecongruent,thenthetrianglesarecongruent.Proof:Wearegiven∆ABCand∆DEFwithAB=DE,∠A=∠D,and∠B=∠E.ByTheorem9,wecansuperposeABontoDEinoneortworeflections.IfC'=F,we’redone.Ifnot,reflectFacrossDEwithimageF'.Sincereflectionspreservesanglemeasure,C'mustbeonrayDF'andonrayEF'.ItfollowsthatC'=F',soareflectionacrossDEsuperposes∆ABConto∆DEF.



12. HLCongruenceCriterionforRightTriangles:Ifthehypotenuseandonelegofonerighttrianglearecongruenttothehypotenuseandonelegofanother,thentherighttrianglesarecongruent.Proof:Wearegiven∆ABCand∆DEFwithhypotenuseAC=hypotenuseDFandlegAB=legDE.ByTheorem9,wecansuperposeABontoDEinoneortworeflections.Because∠ABCisarightangleandreflectionpreservesanglemeasure,B'C'⊥A'B'.But∠DEFisarightanglealsoandreflectionpreservessegmentlength,soC'liesonthelinethroughEperpendiculartoDEandonthecirclewithcenterDandradiusDF.IfC'=F,we’redone.Ifnot,reflectFacrossDEwithimageF'.C'mustlieonrayEF'andonthecirclewithcenterDandradiusDF.ItfollowsthatC'=F',soareflectionacrossDEsuperposes∆ABConto∆DEF.

13. Iftwotrianglesarecongruent,onecanbesuperposedontheotherbyasequenceofatmostthreereflections.Proof:TheproofsinTheorem11and12showthis.

14. *Angleswithequalmeasurearecongruent.IfwecombinethiswithTheorem5,wehave:AnglesarecongruentifandonlyiftheyhaveequalmeasureProof:Given∠A=∠D,construct∆ABCand∆DEFwithAB=AC=DE=DF.∆ABC≅∆DEFbySAS,soasequenceofreflectionssuperimposes∆ABConto∆DEF.Thisimpliesthat∠Amapsonto∠D.

TwoReflections

15. *Ifalinelisperpendiculartooneoftwodistinctparallellineseandf,itisperpendiculartotheother.Proof:Inthediagram,lisperpendiculartoeatPandfintersectslatQ.Assumethatfisnotperpendiculartol.Reflecteandfinl.BecauseQisonl,Q'=Q,andbecauseeisperpendiculartol,e'=e.Iffwerenotperpendiculartol,f'≠f.f'can'talsobeparalleltoe,becausethatwouldcontradicttheparallelpostulate.So,eandf'intersectatapointR.ReflectingagaininlwouldshowthateandfintersectatR'.Butthisisimpossible,becauseeandfareparallel.Therefore,ourassumptionisincorrect,andfmustbeperpendiculartol.



16. Thecompositionoftworeflectionsinparallellinesisatranslation.Thetranslationvectorisperpendiculartothelines,pointsfromthefirstlinetothesecond,andhaslengthtwicethedistancebetweenthelines.Thisimpliesthatanytranslationcanbedecomposedintotworeflections.Proof:Inthediagram,thereflectionlineseandfareparallel.AssumepointAisontheoppositesideofefromfasshown.AreflectstoA'ine;A'reflectstoA"inf.Bythedefinitionofreflection,AA'⊥e,A'A"⊥f,andtheequalsegmentsarelabeledxandy.Sincelinej,whichcontainsAandA',isperpendiculartoe,itisalsoperpendicularto

fbyTheorem15.So,A,A',andA"arecollinear.Therefore,AtranslatestoA"bythevectorshown,whichisperpendiculartoeandf.Itslength,2(x+y),istwicethedistancex+ybetweeneandf.IfAisbetweeneandf,thediagramlookslikethis:Asbefore,thedistancebetweenAandeisxandthedistancebetweenA'andfisy.Nowthedistancebetweeneandfisy–xandthe

distancebetweenAandA"is2y–2x,buttheargumentisessentiallythesame.Thisargumentalsoappliesinthethirdcase,whereAisontheoppositesideofffrome.Thediagramforthiscaseisbelow. Foradynamicviewofthiswithaquadrilateralaspre-image,seeTheorem3onthiswebpage:http://www.mathedpage.org/transformations/isometries/four/index.html#two.



17. Thecompositionoftworeflectionsinintersectinglinesisarotationaroundtheirpointofintersection.Theangleofrotationistwicethedirectedanglebetweenthelinesgoingfromthefirstreflectionlinetothesecond(eitherclockwiseorcounterclockwise).Thisimpliesthatanyrotationcanbedecomposedintotworeflections.Proof:TheargumentislikeTheorem16.Inthediagram,thereflectionlineseandfintersectatOandtheresultingrotationiscounterclockwise.AreflectstoA'ine;A'reflectstoA"inf.Sincereflectionpreservesanglesandthereflectionlineisfixed,theequalanglesarelabeledxandy.Bythedefinitionofrotation,OA=OA'=OA".Oneanglebetweeneandfisx+y,andthecounterclockwiserotationanglefromAtoA"is2x+2y=2(x+y),twicetheanglebetweenthelines.Theotheranglebetweeneandfis180–(x+y)andtheclockwiserotationangleis360–2x–2y=360–2(x+y)=2[180–(x+y)],alsotwicetheanglebetweenthelines.Foradynamicviewofthiswithaquadrilateralaspre-image,seeTheorem4onthiswebpage:http://www.mathedpage.org/transformations/isometries/four/index.html#two.

18. *Rotation,andtranslationpreservesegmentlength,anglemeasure.collinearity,andbetweenness.Proof:Weassumedthatreflectionpreservessegmentlengthandanglemeasure.Theorem2showsthatreflectionpreservescollinearityandbetweenness.Sincerotationandtranslationarecompositionsoftworeflections,theypreservethesealso.

19. *IfA→A'andB→B'underarotationortranslation,segmentABmustmapontosegmentA'B'.Proof:Bothrotationandtranslationcanbedecomposedintotworeflections.Sincethetheoremistrueforreflections(Theorem3),itmustbetrueforrotationandtranslation.

20. *Reflections,rotations,andtranslationsmapraysontoraysandlinesontolines.Proof:Bothrotationandtranslationcanbedecomposedintotworeflections.Sincethetheoremistrueforreflections(Theorem4),itmustbetrueforrotationandtranslation.

21. *Giventwointersectinglines,therearetworeflectionsmappingonetotheother.Proof:Thelinesthatbisecteachoftwoadjacentanglesformedattheintersectionpointarethelinesofreflectionbecausereflectionpreservesanglemeasure.Note:TherelatedtheoremfordistinctparallellinesisTheorem10ofChapter3.

Half-Turns,VerticalAngles,Transversals,andTranslations



22. *IfA'=Bunderahalf-turn,thenB'=A.Proof:180˚rotationclockwiseisequivalentto180˚rotationcounter-clockwise.So,ifahalf-turntakesAtoB,itsinverse(itself)willtakeBtoA.

23. *TheimageofsegmentABunderahalf-turnarounditsmidpointisthesegmentBA.Thatis,A'=BandB'=A.Proof:IfMisthemidpointofsegmentAB,∠AMB=180˚andMA=MB.Bythedefinitionofrotation,A'=BandB'=Aunderahalf-turnaroundM.Sincetheimageofasegmentisasegment,ABisinvariantunderthehalf-turn.

24. *Alineisitsownimageunderahalf-turnaroundapointontheline.Proof:IfOisthecenterofrotationandPisapointontheline,∠POP'=180˚underthehalf-turnaroundO.Therefore,P'isalsoontheline.Underthehalf-turn,(P')'=P,soeverypointonthelineistheimageofanotherpoint.

25. *Theimageofalineunderahalf-turnisparalleltothepre-image.Proof:Theorem23provesthisifthecenterofrotationisontheline,sosupposethatitisn’t.f'istheimageoffandA'istheimageofAunderahalf-turnHaroundP.ByTheorem21,HwilltakeA'backtoA.SincethisistrueforallpointsAonlinef,Hwilltakef'backtof.Supposef'isnotparalleltof,sothatf'andfintersectatQ.QmustbedistinctfromP,becauseitliesonf.LetQ'=H(Q).Q'isanintersectionpointoffandf'thatisdistinctfromQ.Sincef'andfintersectintwodistinctpoints,f'=f.ButifA'isonf,thenPisalso,contradictingourassumption.

26. *Whentwolinesintersect,theverticalanglesareequal.Proof:Rotateangleα180˚aroundP.α'=αbecauserotationpreservesangles.ButlinesfandgaretheirownimagesbyTheorem23.Soα'=βandthereforeα=β.



27. *Iftwodistinctlinesarecutbyatransversal,theyareparallelifandonlyifthealternateinterioranglesareequal.Proof:Supposethelinewecalledf'isparalleltofandthetransversalintersectsfatAandf'atapointwe'llcallA'.LetPbethemidpointofsegmentAA'.LetHbethehalf-turnaroundpointP.ByTheorem22,A'istheimageofAunderH.ByTheorem24,f'istheimageoff.SincePisfixedunderH,segmentPA'istheimageofsegmentPA.Therefore,angleα'istheimageofangleα,andbecauserotationpreservesanglemeasure,α'=α.Conversely,supposeintheseconddiagramthatthatfandgaretwodistinctlines,andthatPisthemidpointoftransversalsegmentAA'.Furthermore,supposethatthealternateinteriorangleslabeledαandα'areequal.SincerotationpreservesanglesandsegmentPA'isonesideofH(α),H(f)=g.ByTheorem24,gisparalleltof.

28. *Iftwodistinctlinesarecutbyatransversal,theyareparallelifandonlyifthecorrespondinganglesareequal.Proof:ByTheorem26,f'isparalleltofifandonlyifα'=α.ByTheorem25,β=α'.Therefore,f'isparalleltofifandonlyifβ=α.



29. *Thecompositionoftranslationsiscommutative.Notation:Tu(B)istheimageofBunderatranslationwithvectoru.Tu(!")istheimageoflineABunderatranslationwithvectoru,andsimilarlywithsegmentsandvectors.Proof:Letuandvbethetwotranslationvectors.Ifthevectorsareparallel,takerepresentativesonthesameline.Markthislinewithnumberstomakeitanumberline.Theresultfollowsfromcommutativityofaddition.Iftheyarenotparallel,letTu(A)=A',Tu(B)=B'andTv(A)=B.WeneedtoshowthatTv(A')=B'.LetTv(A')=C.Bythedefinitionofvector,A'C=AB(thelengthofv).A'B'=ABalso,sinceA'B'=Tu(AB)andtranslationpreservesdistance.SinceA'B'=A'C,bothB'andCareonacirclecenteredatA',withradiusAB.Likewise,BB'=AA'(thelengthofu),andBC=AA'sinceBC=Tv(AA').Therefore,bothB'andCareonacirclecenteredatB,withradiusAA'.Thus,B'andCarebothattheintersectionofthetwocircles.LetmbethelinethroughAandB.Thenm'=Tu(m)isthelinethroughA'andB'.B'mustbeonthesamesideofmasA',becausetheotherintersectionisT-u(B).CmustbeonthesamesideofmasA'becauseTv(A')=Candvisthedirectionofm.ThereforeB'andCarethesameintersectionpoint,soTu(Tv(A)=Tu(B)=B'andTv(Tu(A)=Tv(A')=B'.Pedagogicalnote:Itisessentialtomakeclearthatingeneral,compositionisnotcommutative.Forexample,studentscouldexperimentoninteractivegeometrysoftware.

30. *Thetranslationimageofalineisparalleltotheline.Proof:Part1:Thetranslationimageofalineisthelineitselfifthevectorisparalleltotheline.Proof:Let! = !"bearepresentativeofthetranslationvectorwithAonlinef.IfPisanarbitrarypointonlinef,!!′ = !bythedefinitionoftranslation.P'liesonfbecause!movespointsinthedirectionoftheline.Tolocatethepre-imageofanypointonthelineunder!,apply−!toit.So,theimageofthelineistheentireline.



Part2:Thetranslationimageofalineisadistinctlineparalleltoitspre-imageifthevectorisnotparalleltotheline.Proof:WecanusethesamediagramandsetupasTheorem28.Consideralinemandavectorunotparalleltom.ChoosetwoarbitrarypointsAandBonm.LetTu(A)=A',Tu(B)=B',and!" = v.WehavealreadyshownthatTv(A')=B',andudoesnothavethesamedirectionasv,sothetworepresentations!"and!"ofvareparallelandnotcollinear.Wehaveshownthattheimageofanysegmentonmisaparallelsegmentonm'.Therefore,m'=Tu(m)isparalleltom.

31. *Anyrepresentativeofavectorcanbesuperimposedonanyotherbyatranslation.Proof:Ifuandvaretworepresentativesofthesamevector,translatetheinitialpointofutotheinitialpointofvbytranslationT.Sinceuandvareparallel,sincetranslationmapsanylineintoaparallelline,andsinceuandvhavethesamelengthandpointinthesamedirection,T(u)=v.

32. *Iftwodistinctlinesarecutbyatransversal,anangleononelineisatranslationimageofanangleontheotherifandonlyifthelinesareparallel.a)Iftwoparallellinesarecutbyatransversal,anangleononeparallelisthetranslationimageofanangleontheother.Proof:Givendistinctparallellinesmandnandtransversaltshowninthediagram,translatembyvector!".m'containsQandisparalleltombyTheorem29,Part2.Bytheparallelpostulate,m'=n.t'=tbyTheorem29,Part1.Thereforeα'=β.Sincetranslationpreservesangles,α=β.b)Iftwolinesarecutbyatransversal,andifanangleononelineisthetranslationimageofanangleontheother,thenthelinesareparallel.Proof:Supposeangleβisatranslationimageofangleα,sothatα'=β.Becauseintersectionmustmaptointersection,!"isthetranslationvectorandtisatransversalbecauseitiscommontobothangles.Sinceα'=β,m'=n.ByTheorem29,Part2,m||n.



SumofAngles33. Thesumoftheanglesofatriangleis180˚.Anexteriorangleofatriangleisequalto

thesumoftheremoteinteriorangles.Thereforeitisgreaterthaneitherone.Proof:RotateΔABC180˚aroundM,themidpointofsegmentBC.ByTheorem22,B'=CandC'=B.β'=βbecauserotationpreservesanglesandBA'||ACbyTheorem26.NowtranslateΔABCbyvector!".A,BandB'arecollinearbyTheorem29Part1,andC'=A'fortworeasons:theimageofACisalinethroughBparalleltoACbyTheorem29Part2,andBA'=ACbecausetranslationpreservesdistance.Because∠ABB'=180˚,α'+β'+γ=α+β+γ=180˚.Also,fromthediagram,exterior∠CBB'=α+βandisgreaterthaneitherαorβ.PedagogicalNote:Aninformalversionofthisproofcanbediscussedwithstudentsafteraskingthemtofindatessellationbasedonascalenetriangletile.)

34. Thesumoftheinterioranglesofaquadrilateralis360˚.(Inaconcavequadrilateral,oneinterioranglewillbegreaterthan180˚.Suchanangleiscalledareflexangle.)Proof:Drawadiagonalfromthevertexofareflexangleifthereisone,arbitrarilyotherwise.Thediagonaldividesthequadrilateralintotwotriangles,thesumofwhoseanglesis180˚.Theanglesofbothtriangles,together,makeuptheinterioranglesofthequadrilateral.



Chapter2:SymmetryDefinitionsandProperties-TrianglesandQuadrilaterals

Wedefinespecialtrianglesandquadrilateralsintermsoftheirsymmetries.Webegineachsectionwithasymmetrydefinition,thenuseittodeducethefigure’sproperties.Theproofsaregenerallyeasierthanthetraditionalapproach.Chapter3dealswiththeconverse:wesupplyconditionsandprovethatthefigurehasthedefiningsymmetry.Thisisamoredifficulttask.BytheendofChapter3,wewillhaveshownthatthetraditionaldefinitionsareequivalenttothesymmetrydefinitions.Thegeneraltrapezoidisaspecialcase,becauseithasnosymmetry.Wecanuseatransformationtodefineit,butnotanisometry.Forthatreason,wepostponetheoremsaboutageneraltrapezoiduntilChapter4:BasicsofDilationandSimilarity.Ourdefinitionsareinclusive,whichislogical,andmoreoverconsistentwiththebehaviorofdynamicgeometrysoftware.Forexample,youcandragavertexofadynamicisoscelestrapezoidtomakeitarectangle,soweconsiderarectangleasaspecialisoscelestrapezoid.Similarly,anequilateraltriangleisaspecialisoscelestriangle,arhombusisaspecialkite,andsoon.PedagogicalNote:Studentsmostlikelyarealreadyfamiliarwiththetraditionaldefinitionsofspecialtrianglesandquadrilaterals.ThischaptercanbeprecededwiththeSymmetricPolygonsactivity,wherestudentsaregivensymmetrydefinitions,andaskedtorecognizewhichpolygoncanbedefinedthatway.[insertlinkhere]

1. *Theimageofavertexinalineofsymmetryisalsoavertex.

Proof:Avertexisthecommonendpointoftwosides.Becausecollinearityispreserved,sidesmustmapontosides.So,theimageofavertexmustalsolieontwosides.Apointontwosidesisavertex,soitmustalsobeavertex.

2. IsoscelesTriangle:Atrianglewithatleastonelineofsymmetry.

Theetymologyof“isosceles”,ofcourse,is“equallegs”.Intheschemewepropose,thisisnolongerthedefinition;itisapropertythatmustbeproved.(Seeproperty(b)below.)Properties:

a. Onevertexliesonthelineofsymmetryandtheothertwoareeachother'sreflections.Proof:Becausethereisanoddnumberofvertices,oneofthemmustlieonthelineofsymmetry.



b. Anisoscelestrianglehastwoequalsidesandtwo

equalangles.Proof:Reflectionpreservessidelengthsandanglemeasure.IfvertexAisonthelineofsymmetry,thenAB=ACand∠B=∠C.

c. Theperpendicularbisectorofthethirdsideofanisoscelestrianglebisectsanangleofthetriangle,sothelineofsymmetryisanaltitude,amedian,andaperpendicularbisector.Proof:Bythedefinitionofreflection,thelineofsymmetrylistheperpendicularbisectorofBC.ItalsomustpassthroughA.Sincereflectionpreservesangles,∠DAB=∠DAC,solbisects∠BAC.lisclearlyanaltitude,median,andperpendicularbisector.

3. EquilateralTriangle:Atrianglewithatleasttwolinesofsymmetry.Other(equivalent)definitionsarepossible.Wepreferthisone,asitiseconomical,andfacilitatestheproofofproperties.Notethatonceagain,theetymologydoesnotcorrespondtothedefinition;wemustproveallsidesareequal.(Seeproperty(b)below.)Properties:

a. Anequilateraltrianglehas3-foldrotationalsymmetry.Proof:LetmandnbethesymmetrylinesthroughAandCrespectively.Thecompositionofreflectionsinm,thennmapsthetriangleontoitselfandisarotationaroundtheirintersectionpointD.Callthisrotationr.rmapsAontoB,BontoC,andContoA.Repeatingthisrotationthreetimesgivestheidentitytransformation,sothetrianglehas3-foldrotationalsymmetryaroundtheintersectionpointofitstwolinesofsymmetry.

b. Allsidesofanequilateraltriangleareequalandeachangleis60˚.Proof:Rotationpreservessidelengthsandanglemeasure.Sincethesumoftheanglesinatriangleis180˚,eachangleis60˚.

c. Anequilateraltrianglehasthreeconcurrentlinesofsymmetry.Proof:rmapsAtoBandDtoitself,som',theimageofmunderr,passesthroughBandD.SincemperpendicularlybisectsBC,m'must



perpendicularlybisectCAbecauserotationpreservessegmentlengthandanglemeasure.Thereforem'isathirdlineofsymmetryofΔABC.

d. Eachlineofsymmetryofanequilateraltriangleisanaltitude,amedian,andaperpendicularbisector.Proof:Thetriangleisisoscelesinthreedifferentways.

Thenextfigure,aparallelogram,isdefinedusingrotationalsymmetry,soweneedafewcommentsaboutvocabulary.Ifafigurecoincideswithitselfntimesinonecompleteturnaboutaspecifiedpoint(oftencalleditscenter),wesaythefigurehasn-foldrotationalsymmetry.Forexample,aregularpentagonhas5-foldrotationalsymmetry.Wealsosaythataregularpentagonhas72˚rotationalsymmetry,becausearotationof72˚arounditscenterwillmapitontoitself.Wecouldmakethesamestatementusinganyintegermultipleof72˚,butit’smoredescriptivetousethesmallestpositiveangle.Ingeneral,thesmallestrotationangleforafigurewithn-foldrotationalsymmetryis!"#˚! .4. Parallelogram:Aquadrilateralwith2-fold(180˚)rotationalsymmetry.

Note:Thisfigureandageneraltrapezoidaretheonlyspecialquadrilateralswhosedefinitionsdonotinvolvelinesymmetry.Properties:

a. *Theimageofavertexunderthesymmetryrotationisanoppositevertex.Proof:Letrbethe2-foldrotation.rfollowedbyr(r°r)isa360°rotation,i.e.theidentity.Aswithtriangles,theimageofavertexunderrmustbeavertex.Itsimageunderr°rmustbeitself.Iftheimagewereaconsecutivevertex,thentheimageunderr°rwouldbethenextconsecutivevertex(i.e.theoppositevertex),nottheoriginal.Therefore,theimageistheoppositevertex.

b. Thecenterofthe2-foldrotationisthecommonmidpointofthediagonals.Proof:Adiagonalmustrotateintoitselfbecauseitsendpointsswitch.So,adiagonalmustcontainthecenterofrotation.Thedistancefromthecentertooneendpointmustequalthedistancetotheotherbecauserotationpreservesdistance.Therefore,thecentermustbethemidpointofeitherdiagonal,whichimpliesitisthecommonmidpointofboth.

c. Theoppositesidesofaparallelogramareparallel.Proof:Theimageofalineunderahalf-turnaroundapointnotonthelineisaparallelline.



d. ConsecutiveanglesofaparallelogramaresupplementaryProof:Thisisapropertyofparallellinescutbyatransversal.

e. *Theimageofasideunderrisanoppositeside.Theimageofanangleunderrisanoppositeangle.Proof:Theimagecan'tbeaconsecutivesidebecausethenitwouldn'tbeparalleltothepre-image.Theimageofananglecan'tbeaconsecutiveanglebecausethenoneimagesidewouldn'tbeparalleltoitspre-image.

f. Theoppositesidesandoppositeanglesofaparallelogramareequal.Proof:Rotationpreservessegmentlengthandanglemeasure.

5. Kite:Aquadrilateralwithatleastonelineofsymmetrythroughoppositevertices.(Itispossibletoomit“opposite”fromthedefinitionandprovethatifalineofsymmetrypassesthroughvertices,theymustbeopposite.Formoststudents,thissortofsubtletywouldbecounterproductive.Ontheotherhand,itwouldmakeforaninterestingdiscussion.)Properties:

a. Akitehastwodisjointpairsofconsecutiveequalsidesandonepairofequaloppositeangles.(Weneedtosay“disjoint,”becausethepairscan'thaveacommonside.)Proof:Reflectioninthelineofsymmetrypreservessegmentlengthandanglemeasure.

b. Thelineofsymmetryofakitebisectsapairofoppositeangles.Proof:Reflectionpreservesanglemeasure.

c. Thediagonalofakitethatliesonthelineofsymmetryperpendicularlybisectstheotherdiagonal.Proof:Thesymmetrylineperpendicularlybisectsthesegmentjoiningtheverticesnotonthelinebecausetheyreflectintoeachother.

6. IsoscelesTrapezoid:Aquadrilateralwithalineofsymmetrythoughinteriorpointsofoppositesides.Ifthereisonlyonesuchlineofsymmetry,thesesidesarecalledbases.Inthiscase,theothertwosidesarecalledlegs.Properties:

a. Twoverticesofanisoscelestrapezoidareononesideofthesymmetrylineandtwoareontheother.Proof:Sincereflectionmapsverticestovertices,thefourverticesmustbeevenlysplitoneithersideofthesymmetryline.



b. Thesymmetrylineofanisoscelestrapezoidistheperpendicularbisectorofthebases.Proof:Oneendpointofeachofthesesidesmustreflectintotheother.Areflectionlineperpendicularlybisectsthesegmentjoiningpre-imageandimagepointsifthesepointsarenotonthereflectionline.

c. Thebasesofanisoscelestrapezoidareparallel.Proof:Theyarebothperpendiculartothesymmetryline.Twodistinctlinesperpendiculartothesamelineareparallel.

d. Thelegsofanisoscelestrapezoidareequal.Proof:Reflectionpreservessegmentlength.

e. Twoconsecutiveanglesofanisoscelestriangleonthesamebaseareequal.Proof:Reflectionpreservesanglemeasure.

f. Thediagonalsofanisoscelestrapezoidareequal.Proof:Onediagonalreflectstotheother.Reflectionpreservessegmentlength.

g. Theintersectionpointoftheequaldiagonalsofanisoscelestrapezoidliesonthesymmetryline.Proof:Thepointwhereonediagonalintersectsthesymmetrylinemustbeinvariantunderreflectioninthesymmetrylinebecauseitliesonit.Therefore,italsoliesontheotherdiagonal.

h. Theintersectionpointofthediagonalsofanisoscelestrapezoiddivideseachdiagonalintoequalsubsections.Proof:Thesubsectionsofonediagonaldeterminedbytheintersectionpointreflectontothesubsectionsoftheother.Thesesubsectionsareequalbecausereflectionpreservessegmentlength.

7. Rhombus:Aquadrilateralwithtwolinesofsymmetrypassingthroughoppositevertices.(So,arhombusisakiteintwodifferentways.)Properties:

a. Arhombushasallsidesequalandtwopairsofequaloppositeangles.Proof:Akitehastwodisjointpairsofconsecutiveequalsidesandonepairofequaloppositeangles.Theresultfollowsbecausearhombusisakiteintwodifferentways(i.e.bothdiagonalsarelinesofsymmetry).

b. Eachdiagonalofarhombusbisectsitsangles.Proof:Eachlineofsymmetrybisectsapairofoppositeangles(propertyofkites).



c. Thediagonalsofarhombusperpendicularlybisecteachother.Proof:Thediagonalofakitethatliesonthelineofsymmetryperpendicularlybisectstheotherdiagonal.Forarhombus,eachdiagonalhasthisproperty.

d. Arhombusisaspecialparallelogram.Proof:Sincearhombushastwoperpendicularlinesofsymmetry,thecompositionofreflectionintheselinesisa180˚rotationaroundtheirpointofintersection.(Thecompositionoftworeflectionsisarotationaroundtheirpointofintersectionthroughtwicetheanglebetweenthereflectionlines.)Sinceeachreflectionmapstherhombusontoitself,theircompositiondoesalso.

e. Theoppositesidesofarhombusareparallel.Proof:Sincearhombusisaparallelogram,theoppositesidesareparallel.

8. Rectangle:Aquadrilateralwithtwolinesofsymmetrypassingthroughinteriorpointsoftheoppositesides.(So,arectangleisanisoscelestrapezoidintwodifferentways.)Properties:

a. Thesymmetrylinesofarectangleperpendicularlybisecttheoppositesides.Proof:Arectangleisanisoscelestrapezoidintwodifferentways.

b. Arectangleisequiangular.Proof:Twoconsecutiveanglesofanisoscelestrapezoidthatshareabaseareequal.Bothpairsofoppositesidesarebasesbecauseofthetwodifferentways,soanytwoconsecutiveanglesshareabase.

c. Allanglesofarectanglearerightangles.Proof:Thesumoftheinterioranglesofanyquadrilateralis360˚and360÷4=90.

d. Thesymmetrylinesofarectangleareperpendicular.Proof:Thelinesdividetherectangleintofourquadrilaterals.Eachhasthreerightangles:oneisanangleoftherectangleandtheothertwoareformedbyasideandasymmetryline,whichareperpendicular.Sincethesumoftheanglesofaquadrilateralis360˚,thefourthangleattheintersectionofthesymmetrylinesmustalsobearightangle.



e. Arectanglehas2-foldrotationalsymmetry,soitisaspecialparallelogram.Proof:Reflectingarectangleinonelineofsymmetryfollowedbytheothermapstherectangleontoitselfandisequivalenttoa180˚rotationbecausethesymmetrylinesmeetatrightangles.Therefore,arectanglehas2-foldrotationalsymmetryaroundtheintersectionofthesymmetrylines.

f. Theoppositesidesofarectangleareparallelandequal.Proof:Thesearepropertiesofaparallelogram.Arectangleisaspecialparallelogram.

g. Thediagonalsofarectangleareequal.Proof:Thisisapropertyofanisoscelestrapezoid.Arectangleisaspecialisoscelestrapezoid.

h. Thediagonalsofarectanglebisecteachother.Proof:Thisisapropertyofaparallelogram.Arectangleisaspecialparallelogram.

i. Thediagonalsofarectangleandthelinesofsymmetryareallconcurrent.Proof:Theintersectionpointoftheequaldiagonalsofanisoscelestrapezoidliesonthesymmetryline.Forarectangle,theintersectionpointliesonbothsymmetrylines,soitistheirintersection.

9. Square:Aquadrilateralwithfourlinesofsymmetry:twodiagonalsandtwolinespassingthroughinteriorpointsofoppositesides.Properties:

a. Asquareisaspecialrectangle,rhombus,kite,andisoscelestrapezoid,soitinheritsallthepropertiesofthesequadrilaterals.Proof:True,bythedefinitionofasquare.

b. Ifasquareandallfoursymmetrylinesaredrawn,alltheacuteanglesare45˚.Proof:Thediagonalsbisecttheinteriorrightanglesbecauseasquareisarhombus.Alleightrighttrianglesformedhavearightanglewherethesymmetrylinesintersectthesidesanda45˚anglewheretheyintersectthevertices.Sincethesumoftheanglesofatriangleis180˚,theremaininganglesatthecentermustallbe45˚.

c. Asymmetrylinethroughsidesandasymmetrylinethroughverticesforma45˚angle.Proof:Animmediateconsequenceoftheresultjustabove.



d. Asquarehas4-foldrotationalsymmetry.Proof:Reflectingasquareinalineofsymmetrythroughthesidesfollowedbyalineofsymmetrythroughtheverticesmapsthesquareontoitself.Itisequivalenttoa90˚rotationbecausethesesymmetrylinesmeetata45˚angle.Therefore,asquarehas4-foldrotationalsymmetryarounditscenter(theintersectionpointofthelinesofsymmetry).



Chapter3:ProvingTrianglesandQuadrilateralsSatisfySymmetryDefinitions

1. IsoscelesTriangle:Atrianglewithonelineofsymmetry.

a. Ifatrianglehastwoequalsides,itisisosceles.Proof:LetABandACbetheequalsides.AmustlieontheperpendicularbisectorlofBCbecauseitisequidistantfromBandC.Bythedefinitionofreflection,B'=Cunderreflectioninl.A'=Abecauseitliesonl.Therefore,lisalineofsymmetryforΔABC.

b. Ifatrianglehastwoequalangles,itisisosceles.Proof1:ByTheorem21ofChapter1,theraysABandACarereflectionsofeachotherintheanglebisectorADof∠BAC.Twoanglesof∆BADand∆CADareequal,sothethirdangles(∠BDAand∠CDA)mustbeequal.Sincetheyaresupplementary,theyarebothrightangles.ItfollowsthatthereflectionofBinrayADmustbeC.SinceAisitsownreflectioninlineAD,ADisalineofsymmetryforthetriangle.Proof2:DrawsegmentEF,whichperpendicularlybisectssideBC.BandCinterchangeunderthisreflection.Becausereflectionpreservesanglesand∠B=∠C,rayBAandrayCAinterchangealso.Aisonbothrays,soA'mustbeonbothraysalso.ButAistheonlypointonbothrays,soA'=Aunderthisreflection.SinceAisafixedpoint,itliesonthereflectionline.Therefore,lineEFisalineofsymmetryforthetriangle.



c. Ifananglebisectorofatriangleisalsoanaltitude,thetriangleisisosceles.Proof:Letl,thebisectorof∠BACbeperpendiculartosideBCatD,sothat∠DAB=∠DACand∠ADB=∠ADC.Twoanglesof∆BADand∆CADareequal,sothethirdangles(∠Band∠C)mustbeequal.ByTheorem1babove,thetriangleisisosceles.

d. Ifanaltitudeofatriangleisalsoamedian,thetriangleisisosceles.Proof:SincealtitudeADisalsoamedian,ADistheperpendicularbisectorofBC.Sinceanypointontheperpendicularbisectorofasegmentisequidistantfromtheendpoints,AB=AC.ByTheorem1aabove,thetriangleisisosceles.

e. Ifananglebisectorofatriangleisalsoamedian,thetriangleisisosceles.Proof:Postponeduntiltheendoftherhombussection(afterTheorem6e),becausearhombusisconstructedintheproof.

2. EquilateralTriangle:Atrianglewithtwolinesofsymmetry.a. Ifatrianglehasallsidesequal,thenit'san

equilateraltriangle.Proof:InΔABC,AB=BC=CA.SinceAB=AC,thetriangleisisosceleswithsymmetrylinem.SinceCA=CB,thetriangleisisosceleswithsymmetrylinen.Sinceithastwolinesofsymmetry,itisequilateral.

b. Ifatrianglehasallanglesequal,thenit'sanequilateraltriangle.Proof:Theargumentisvirtuallyidenticaltothepreviousone,butusesTheorem1binsteadof1a.

3. Parallelogram:Aquadrilateralwith2-foldrotationalsymmetry.a. Ifthediagonalsofaquadrilateralbisecteachother,thequadrilateralisa

parallelogram.Proof:RotatequadrilateralABCD180˚aroundpointE,theintersectionofthediagonals.Sincetherotationis180˚,B'liesonrayED.Sincerotationpreservesdistance,B'=D.Similarly,A'=C.Similarly,C'=AandD'=B.Becauserotationmapssegmentstosegments,eachsideofABCDmapstotheoppositeside.Therefore,quadrilateralABCDhas2-foldrotationalsymmetry.Bydefinition,ABCDisaparallelogram.



b. Ifoppositesidesofaquadrilateralareparallel,thequadrilateralisaparallelogram.Proof:GivenquadrilateralABCDasinthisfigure,withsidesextendedtolinesk,l,m,andn.Wewouldliketoprovethatithas2-fold(half-turn)symmetry.

LetMbethemidpointofdiagonalAC.

ConsiderH,thehalf-turnwithcenterM.SinceMisthemidpointofsegmentAC,A'=CandC'=AunderH.BecauseMisonneitherknorl,theirimagelinesareparalleltotheirpre-images.Becauseoftheparallelpostulate,thereisonlyoneparalleltokthroughCandoneparalleltolthroughA.Therefore,k'=m,andl'=n.Bistheintersectionoflineskandl,andthereforeitsimageistheintersectionoflinesmandn,whichisD.SinceA'=CandB'=D,ABCDhas2-foldrotationalsymmetry.Bydefinition,ABCDisaparallelogram.



c. Ifoppositesidesofaquadrilateralareequal,thequadrilateralisaparallelogram.Proof:InquadrilateralABCD,drawdiagonalACanditsmidpointE.UnderahalfturnaroundE,A'=CandC'=A.SinceCB=AD,B'liesoncircleAwithradiusAD.SinceAB=CD,B'liesoncircleCwithradiusCD.ThesecirclesintersectatDandF.ButFisonthesamesideoflineACasB,soB'≠F.ThereforeB'=DandABCDhas2-foldsymmetryaroundE.Bydefinition,ABCDisaparallelogram.

d. Iftwosidesofaquadrilateralareequalandparallel,thequadrilateralisaparallelogram.Proof:InquadrilateralABCD,supposeABisequalandparalleltoDC.DrawdiagonalBDanditsmidpointM.Underahalf-turnaroundM,B'=DandD'=B.TheimageofrayBAisparalleltoAB,soitmustcoincidewithrayDC.BecauseAB=DC,thatmeansthatA'=C,andthereforeC'=A.HenceABCDisaparallelogram.

4. Kite:Aquadrilateralwithonelineofsymmetrythroughoppositevertices.a. Iftwodisjointpairsofconsecutivesidesofaquadrilateralareequal,the

quadrilateralisakite.Proof:InquadrilateralABCD,supposeAB=ADandCB=CD.SinceAandCarebothequidistantfromBandD,theylieintheperpendicularbisectorofdiagonalBD.Therefore,listheperpendicularbisectorofdiagonalBD.Underreflectioninl,B'=DandD'=B.BecauseAandCbothlieonl,A'=AandC'=C.listhereforealineofsymmetryandABCDisakite.

b. Ifadiagonalofaquadrilateralbisectsapairofoppositeangles,thequadrilateralisakite.Proof:LabelaslthelinethroughdiagonalACthatbisects∠BADand∠BCD.Considerreflectioninl.SinceAandCareonl,A'=AandC'=C.Since∠BAC=∠DAC,B'liesonrayAD.Since∠BCA=∠DCA,B'liesonrayCD.BecausetheseraysintersectatD,B'=D,whichimpliesthatD'=B.Therefore,lisalineofsymmetryandABCDisakite.



c. Ifonediagonalofaquadrilateralperpendicularlybisectstheother,thequadrilateralisakite.Proof:InquadrilateralABCD,diagonalACperpendicularlybisectsdiagonalBD.LetlbethelinethroughAandC.ReflectABCDinl.SinceAandClieonl,A'=AandC'=C.Bythedefinitionofreflection,B'=DandD'=B.Therefore,lisalineofsymmetryandABCDisakite.

5. IsoscelesTrapezoid:Aquadrilateralwithalineofsymmetrythoughinteriorpointsofoppositesides.

a. Ifonepairofoppositesidesofaquadrilateralareparallelandapairofconsecutiveanglesononeofthesesidesareequal,thequadrilateralisanisoscelestrapezoid.Proof:InquadrilateralABCD,DC||ABand∠A=∠B.LetlbetheperpendicularbisectorofAB.Underreflectioninl,A'=BandB'=A.SinceDC||AB,DC⊥l.Therefore,D'liesonrayDC.BecauseA'=B,rayABmapstorayBA,∠A=∠B,andreflectionpreservesangles,D'liesonrayBC.TheseraysintersectatC,soD'=C.Thus,lisalineofsymmetryforABCDandABCDisanisoscelestrapezoid.

b. Iftwodisjointpairsofconsecutiveanglesofaquadrilateralareequal,thequadrilateralisanisoscelestrapezoid.Proof:InquadrilateralABCD,∠B=∠Aand∠C=∠D.Because∠A+∠B+∠C+∠D=360˚,2∠A+2∠D=360˚.Dividingbothsidesby2gives∠A+∠D=180˚,soDC||AB.ByTheorem5a,ABCDisanisoscelestrapezoid.



c. Iftwooppositesidesofaquadrilateralareparallelandiftheothertwosidesareequalbutnotparallel,thenthequadrilateralisanisoscelestrapezoid.Proof:InquadrilateralABCD,DC||AB,AD=BC,andADisnotparalleltoBC.ThroughB,drawalineparalleltoADmeetingrayDCatE.SinceABDEhastwopairsofoppositeparallelsides,itisaparallelogram.Becausetheoppositeanglesofaparallelogramareequal,∠A=∠BEC.Theoppositesidesofaparallelogramarealsoequal,soAD=BC=BE.Iftwosidesofatriangleareequal,thetriangleisisosceles,whichimpliesthat∠BEC=∠BCE.Finally,becauseDC||AB,∠BCE=∠ABC.Thechainofequalanglesnowreads∠A=∠BEC=∠BCE=∠ABC.ThismeansthatABCDhasapairofconsecutiveequalanglesononeofitsparallelsides.ByTheorem5a,ABCDisanisoscelestrapezoid.

d. Ifalineperpendicularlybisectstwosidesofaquadrilateral,thequadrilateralisanisoscelestrapezoid.Proof:Thetwosidescan’tbeconsecutive,becauseiftheywere,youwouldhavetwoconsecutiveparallelsides,whichisimpossible.InquadrilateralABCD,listheperpendicularbisectorofABandDC.Underreflectioninl,therefore,D'=CandA'=B.Thus,lisalineofsymmetryandABCDisanisoscelestrapezoid.



e. Iftwosidesofaquadrilateralareparallel,andifthediagonalsareequal,thequadrilateralisanisoscelestrapezoid.Proof:InquadriInquadrilateralABCD,DC||ABandAC=BD.LetMbethemidpointofBC.RotateABCD180˚aroundM.SinceB'=C,C'=B,BD'||DC,andCA'||AB,A'isonrayDCandD'isonrayAB.Becauserotationpreservessegmentlength,BD=CD'.Therefore,AC=BD=CD'.ConsiderΔACD'.Sincetwosidesareequal,itisisosceles,so∠BAC=∠BD'C.BDisalsorotated180˚aroundM,soD'C||BD.UsingtransversalAD',weseethat∠BD'C=∠ABD.Thus∠BAC=∠BD'C=∠ABD.BecausetwoanglesinΔABEareequal,thetriangleisisosceles,whichimpliesthatAE=BE.Inotherwords,EisequidistantfromAandB,soitmustlieontheperpendicularbisectorofAB.AsimilarargumentshowsthatEliesontheperpendicularbisectorofDA.SincetheperpendicularbisectorsofABandDCpassthroughthesamepointE,theycoincide.ByTheorem5d,ABCDisanisoscelestrapezoid.

6. Rhombus:Aquadrilateralwithtwolinesofsymmetrypassingthroughoppositevertices.(So,arhombusisakiteintwodifferentways.)

a. Ifthediagonalsofaquadrilateralperpendicularlybisecteachother,thequadrilateralisarhombus.Proof:Bythedefinitionofreflection,thetwoverticesnotoneitherdiagonalareimagesofeachotherunderreflectioninthatdiagonal.Therefore,bothdiagonalsarelinesofsymmetry,whichisthedefinitionofarhombus.

b. Ifaquadrilateralisequilateral,itisarhombus.Proof:Sinceoppositesidesareequal,thequadrilateralisaparallelogram.Therefore,thediagonalsbisecteachother.Sincetwodisjointpairsofconsecutivesidesareequal,thequadrilateralisakite.Therefore,thediagonalsareperpendicular.Nowweknowthatthediagonalsperpendicularlybisecteachother,sothequadrilateralisarhombusbyTheorem6a.



c. Ifbothdiagonalsofaquadrilateralbisectapairofoppositeangles,thequadrilateralisarhombus.Proof:Considerthediagonalsseparately.Since∠ABCand∠ADCarebisected,ABCDisakitewithAD=CDandAB=CB.Since∠BADand∠BCDarebisected,ABCDisakitewithAD=ABandCD=CB.Therefore,allfoursidesareequalandthequadrilateralisarhombusbyTheorem6b.

d. Ifbothpairsofoppositesidesofaquadrilateralareparallel,andiftwoconsecutivesidesareequal,thequadrilateralisarhombus.Proof:Sincebothpairsofoppositesidesareparallel,thequadrilateralisaparallelogram.Therefore,bothpairsofoppositesidesareequal.Sinceapairofconsecutivesidesareequal,allfoursidesmustbeequal.HencethequadrilateralisarhombusbyTheorem6b.

e. Ifadiagonalofaparallelogrambisectsanangle,theparallelogramisarhombus.Proof:InparallelogramABCD,diagonalACbisects∠DAB.Theoppositesidesofaparallelogramareparallel,sothisimpliesthat∠DCBisbisectedaswellbyanglepropertiesofparallellines.ByTheorem4b,ABCDisakite.Therefore,ABCDisarhombusbyTheorem6d.

NowwearereadytoproveTheorem1e:Ifananglebisectorofatriangleisalsoamedian,thetriangleisisosceles.Proof:InΔABC,rayADbisects∠BACandBD=CD.RotateΔABC180˚aroundD.BecauseDisthemidpointofBC,B'=DandD'=B.Becauserotationpreservesdistance,AD=A'D.NowthediagonalsofquadrilateralABA'Cbisecteachother,soABA'Cisaparallelogram.ButdiagonalAA'bisects∠BAC,sobyTheorem6e,ABA'Cisarhombus.Arhombusisequilateral,soAB=AC.



7. Rectangle:Aquadrilateralwithtwolinesofsymmetrypassingthroughmidpointsoftheoppositesides.(So,arectangleisanisoscelestrapezoidintwodifferentways.)

a. Ifaquadrilateralisequiangular,itisarectangle.Proof:Because∠A=∠Band∠C=∠D,ABCDisanisoscelestrapezoidwithlineofsymmetrythroughmidpointsofABandDCbyTheorem5b.Similarly,∠A=∠Dand∠B=∠C,soABCDisanisoscelestrapezoidwithlineofsymmetrythroughmidpointsofADandBC.Bydefinition,ABCDisarectangle.

b. Ifaparallelogramhasarightangle,thentheparallelogramisarectangle.Proof:Suppose∠A=90˚.Then∠C=90˚becauseoppositeanglesofaparallelogramareequal.Thesumoftheinterioranglesofaquadrilateralis360˚,whichleavesatotalof180˚for∠Band∠D.Sincetheyarealsoequal,theymustberightanglesaswell.HenceallanglesareequalrightanglesandthequadrilateralisarectanglebyTheorem7a.

c. Anisoscelestrapezoidwitharightangleisarectangle.Proof:SupposeABCDisanisoscelestrapezoidwithlineofsymmetrypassingthroughbasesABandDC.Withoutlossofgenerality,wecansupposethat∠A=90˚.Becausethebasesofanisoscelestrapezoidareparallel∠EDC=∠A=90˚.∠ADCand∠EDCaresupplementary,so∠ADC=90˚also.Wealsoknowthattwoconsecutiveanglesofanisoscelestrapezoidonthesamebaseareequal,so∠B=∠A=90˚and∠C=∠ADC=90˚.NowABCDisequiangular,sobyTheorem7aitisarectangle.



d. Ifthediagonalsofaparallelogramareequal,theparallelogramisarectangle.Proof:BecauseABCDisaparallelogram,AB||DCanddiagonalsACandBDbisecteachother.Sincethediagonalsareequalaswell,ABCDisanisoscelestrapezoidwhoselineofsymmetrypassesthroughmidpointsofABandDCbyTheorem5e.BythesameargumentwithparallelsidesADandBC,ABCDisanisoscelestrapezoidwhoselineofsymmetrypassesthroughmidpointsofADandBC.ItfollowsthatABCDsatisfiesthesymmetrydefinitionofarectangle.

8. Square:Aquadrilateralwithfourlinesofsymmetry:twopassingthroughoppositeverticesandtwopassingthroughinteriorpointsofoppositesides.

a. Arectanglewithconsecutiveequalsidesisasquare.Proof:Arectanglehastwolinesofsymmetrypassingthroughtheoppositesides.Theoppositesidesofarectangleareequal,soiftwoconsecutivesidesarealsoequal,itisequilateral.Anequilateralquadrilateralisarhombus,soitsdiagonalsareadditionallinesofsymmetry.Therefore,therectangleisasquare.

b. Arhombuswithconsecutiveequalanglesisasquare.Proof:Thediagonalsofarhombusarelinesofsymmetry.Theoppositeanglesofarhombusareequal,soiftwoconsecutiveanglesarealsoequal,itisequiangular.AnequiangularquadrilateralisarectanglebyTheorem7a,soithastwoadditionallinesofsymmetrypassingthroughoppositesides.Therefore,therhombusisasquare

c. Anequilateralquadrilateralwitharightangleisasquare.Proof:AnequilateralquadrilateralisarhombusbyTheorem6b.Oppositeanglesofaquadrilateralareequalandthesumoftheanglesis360˚,soallanglesarerightanglesandthequadrilateralisalsoequiangular.AnequiangularquadrilateralisarectanglebyTheorem7a.Ifaquadrilateralisbotharhombusandarectangle,ithasfourlinesofsymmetryandisthereforeasquare.

d. Anequiangularquadrilateralwithconsecutiveequalsidesisasquare.Proof:AnequiangularquadrilateralisarectanglebyTheorem7a.Theoppositesidesofarectangleareequal,andifconsecutivesidesarealsoequal,itmustbeequilateral.AnequilateralquadrilateralisarhombusbyTheorem6b.Ifaquadrilateralisbotharhombusandarectangle,ithasfourlinesofsymmetryandisthereforeasquare.



e. Aquadrilateralwith4-foldrotationalsymmetryisasquare.Proof:Sinceaquadrilateralhasfoursides,consecutivesidesandanglesmustmaptoeachotherundera90˚rotation.Becauserotationspreservesidesandangles,thequadrilateralmustbebothequilateralandequiangular,whichimpliesthatitisbotharectangleandarhombus.Ifaquadrilateralisbotharhombusandarectangle,ithasfourlinesofsymmetryandisthereforeasquare.

9. AdditionalTriangleTheoremsa. Themediantothehypotenuseofaright

trianglehashalfthelengthofthehypotenuse.Proof:InrighttriangleABC,BD=CD.RotateΔABCandmedianAD180˚aroundD.Becauserotationspreservesegmentlengthandtherotationis180˚,DisthemidpointofAA'aswellasBC.BecausethediagonalsofquadrilateralABA'Cbisecteachother,itisaparallelogrambyTheorem3a.ButaparallelogramwitharightangleisarectanglebyTheorem7b,andthediagonalsofarectangleareequal.ThusAD=!!!!

! = !!!".

b. Asegmentjoiningthemidpointsoftwosidesofatriangle(calledamidsegment)isparalleltothethirdsideandhalfaslong.Proof:IntriangleABC,DandEaremidpointsofACandBCrespectively.RotateΔABCandsegmentDE180˚aroundpointE.SinceEisamidpoint,B'=CandC'=B.Therefore,quadrilateralABA'Chas2-foldrotationalsymmetry,sobydefinition,itisaparallelogram.BecauserotationpreservessegmentlengthandDisamidpoint,AD=DC=D'B.ButADisparalleltoBD'aswell,soABD'DisalsoaparallelogrambyTheorem3d.Theoppositesidesofaparallelogramareparallel,soDE||AB.Becauserotationpreserveslength,DE=ED'=!!!!',andbecausetheoppositesidesofaparallelogramareequal,DD'=AB.HenceDE=!!!".Note:Theproofisshorterandmoreelegantusingdilation.(SeeChapter4)



10. Giventwodistinctparallellinesmandn,thereisareflectionmappingonetotheother.Proof:DrawlinefperpendiculartomatPintersectingnatQ.fisalsoperpendiculartonbyTheorem15ofChapter1.(Alineperpendiculartooneoftwodistinctparallellinesisperpendiculartotheother.)Drawr,theperpendicularbisectorofsegmentPQ.Underreflectioninr,P'=Qbydefinition.WewillshowthatanyotherpointRonlinemreflectsinrtoapointonlinen.ThroughR,drawlinegperpendiculartom;gwillalsobeperpendicularton.Sincecorrespondinganglesareequal,gisparalleltof.ByTheorem15again,gisperpendiculartor.ABRPandQSABareparallelogramswitharightangle,sobyTheorem7btheyarerectangles.Chapter2,Theorem8fsaysthattheoppositesidesofarectangleareequal,soRB=AP=QA=SB.SinceristheperpendicularbisectorofsegmentRS,RreflectsintoS,whichisonlinen.Corollary:Parallellinesareeverywhereequidistant.Proof:PQ=RSsincePA+AQ=RB+BS.



Chapter4:DilationandSimilarityTobeginthischapter,weexpandonthefoundationalideasofChapter1–BasicsofIsometriesandCongruence.Definitions

Dilation:AdilationwithcenterOandscalefactork≠0mapsOtoitselfandanyotherpointPtoP'sothatO,P,andP'arecollinearandthedirectedsegmentOP'=k·OP.(Ifkisnegative,raysOP'andOPpointinoppositedirections.)Similarity:Twofiguresaresimilarifonecanbesuperposedontheotherbyadilationfollowedbyasequenceofisometries.GeneralTrapezoid:Aquadrilateralwhereonesideisadilationortranslationoftheoppositeside.Ifatranslation,thetrapezoidisaparallelogrambecauseonepairofoppositesidesisbothparallelandequal.(Underourinclusivedefinitions,aparallelogramisaspecialtrapezoid.)Note:ThepropertiesofageneraltrapezoidareprovedinTheorem15.ThesepropertieswerenotlistedinChapter2becausethedefinitionofageneraltrapezoiddependsondilation.

Postulates

Toourpreviouslist(theparallelpostulate,reflectionisanisometry,andtheconstructionpostulates),weaddonenecessaryassumptionaboutdilation:6. Dilationpreservescollinearity.

BasicTheoremsAsinChapter1,weuseanasterisktoindicatetheoremsthatinourviewshouldbediscussedinformally,ratherthanprovedformallywhenworkingwithstudents.

1. *Adilationwithscalefactor-1isahalf-turnaroundthecenterofdilation.Proof:Thisfollowsimmediatelyfromthedefinitionsofdilationandrotation.

2. *Adilationwithscalefactork<0isthecompositionofadilationwithscalefactor-1andadilationwithscalefactor|k|,allwiththesamecenter.Proof:Thisisanotherimmediateconsequenceofthedefinitions.

3. *IfOisthecenterofadilation,kisthescalefactor,andO,A,andBaredistinctcollinearpoints,thenA'B'=|k|AB.Proof:Case1:AandBareonthesamesideofO.Bythedefinitionofdilationandthegiveninformation,O,A,A',B,andB'areallcollinear.Evenifkisnegative,A',andB'areonthesamesideofO.Suppose,withoutlossofgenerality,thatAisbetweenOandB,sothatOA+AB=OB.HenceOA<OBand,evenifkisnegative,OA'=|k|OA<|k|OB=OB'.Therefore,A'isbetweenOandB'andOA'+A'B'=OB'.Bythedefinitionofdilation,OB'=|k|OBandOA'=|k|OA.Therefore,A'B'=OB'–OA'=|k|OB–|k|OA=|k|(OB–OA)=|k|AB.Case2:AandBareonoppositesidesofO.Theargumentisverysimilar,exceptthatonestartswithAO+OB=AB.It'sstillthe



casethatO,A,A',B,andB'areallcollinear,butnowOisbetweenA'andB'.Case3:EitherAorBisthesameasO.Inthiscase,A'B'=|k|ABbythedefinitionofdilation.

4. *Theimageofalineunderadilationisaline.Proof:LetObethecenterofadilation.Ifthepre-imagelinemcontainsO,thentheimagewillbethesamelinewithpointsmapped|k|timestheirdistancefromO,onthesamesideifk>0andontheothersideifk<0.So,m'=minthiscase.NowsupposemdoesnotcontainO.Sincedilationspreservecollinearity,m'willbeapossiblypropersubsetofaline.Neithermnorthelinecontainingm'canbeparalleltoOP.PickanypointQonthelinecontainingm'anddrawlineOQ.Becauseneithermnorm'isparalleltolineOQ,itwillintersectmatP.SincethedilationmapsPtoQ,anypointonthelinecontainingm'isanimagepoint.Therefore,theimageofmisanentireline.

5. FundamentalTheoremofDilations(FTD):IfC,A,andBarenotcollinear,thesegmentA'B'joiningtheimagesofAandBunderadilationwithcenterCandscalefactorkisparalleltosegmentABandhaslength|k|AB.Proof:Part1:SegmentA'B'||segmentAB.Proof:LettheimageoflinemthroughpointsAandBbem'.Assumethatmandm’arenotparallel.ThentheymeetatapointB≠C.ChooseanotherpointAonm.TheimageofAistheintersectionoflineCAwithlinem'.Bisitsownimage,soB'=B.FromAandA',k≠1.FromBandB',k=1.Thiscontradictioncompletestheproof.Part2:Ifk>0,A'B'=kABandthedirectedsegmentsA'B'andABpointinthesamedirection.Proof:LetA'andB'betheimagesofAandBrespectively,withthelengthsofsegmentsCA,CA',CB,andCB'aslabeledinthefigure.Fromthedefinitionofdilation,! = !!!

! = !!!! .ByPart1,ABandA'B'are

parallel.DrawalinethroughBparalleltoAA'intersectingA'B'atK.Nowdilate∆ABCfromcenterBwithscalefactor− !

! ,resultinginthefigureshown.Weindicatesegmentlengthsobtainedfromknowingthatoppositeparallelsidesofaquadrilateralinsurethatit'saparallelogram,andtheoppositesidesofa

parallelogramareequal.Bythedefinitionofdilation,!! =!!.Therefore,

1+ !! = 1+ !

!,or!!!! = !!!

! .Intermsofthesegments,thistranslatesto!!!!" =

!!!!!" =k,soA'B'=kAB,asdesired.



Part3:Ifk<0,A'B'=|k|ABandthedirectedsegmentsA'B'andABpointinoppositedirections.Proof:ByTheorem2,wecandecomposethedilationtoacompositionofahalf-turnaroundthecenterfollowedbyadilationfromthecenterwithscalefactor|k|.TheimageofthedirectedsegmentABunderahalfturnhasthesamelengthandpointsintheoppositedirection.NowweapplyPart2abovetoobtaintheresult.Pedagogicalnote:Theproofischallenging.Ifityoudeemittobetoomuchforyourstudents,youcanconsiderthistheoremtobeanaxiom,asitisintheCommonCoreStateStandards.Wepresentitasatheorembecause,althoughitdoesnotfollowfromthedefinitionofdilation,itcanbederivedfromtheassumptionthatdilationspreservecollinearity.

6. Underadilation,A'B'=|k|AB.ThedirectedsegmentsABandA'B'pointinthesamedirectionifk>0andinoppositedirectionsifk<0.Proof:Theorem3provesthisifO,A,andBarecollinear.Theorem5(theFTD)provesitifnot.

7. *Dilationpreservesbetweenness.Proof:IfpointsA,BandCareinorderonaline,AB+BC=AC.Ifkisthescalefactorofthedilation,thenA'B'=|k|AB,B'C'=|k|BC,andA'C'=|k|ACbyTheorem5.Therefore,A'B'+B'C'=A'C'.ThisimpliesthatB'isbetweenA'andC'.

8. *Theimageofasegmentunderadilationisasegmentandtheimageofarayisaray.Proof:LetAandBbetwopointswithimagesA'andB'underadilationwithcenterOandscalefactork.Becausecollinearityandbetweennessarepreserved,theimageofsegmentABisasubsetsegmentA'B',andsimilarlyforrays.AnypointQonsegmentA'B'orrayA'B'istheimageofapointQ'underadilationwithcenterOandscalefactor!!.Becausecollinearityandbetweennessarepreserved,Q'mustlieonsegmentABorrayAB.So,theimageistheentiresegmentorray.

9. Dilationpreservesanglemeasure.Proof:Theimageofanyrayunderadilationisanotherraythatiseitherparallelto,orcollinearwith,itspre-image.(Ifk<0,thedirectionsofbothrayswillbereversed.)Thesameistrueforatranslationandahalf-turn(neededifk<0).So,withatranslationandahalf-turnifk<0,wecanmapanyangleintoitsimageunderadilation.Sincetranslationsandhalf-turnspreserveangles,dilationsdotoo.

10. Dilationpreservestheratioofthelengthsofanytwosegments.Proof:LetABandCDbetwosegments.A'B'=|k|ABandC'D'=|k|CDbyTheorem6.Therefore,!

!!!!!!! =

!"!".

SimilarTriangles

11. Similartriangleshavecongruentanglesandproportionalsides.Proof:DilationpreservesanglemeasurebyTheorem9andisometriesdoalso.DilationpreservestheratioofsidelengthsbyTheorem10andisometriesdoalsobecausetheypreservedistance.



12. SimilarityCriteriaforTrianglesa. SSSSimilarity:Ifthesidesoftwotrianglesareproportional,thenthe

trianglesaresimilar.Proof:Assume∆ABCand∆DEFhaveproportionalsides,withratiokasshowninthediagram.Dilate∆ABCwithanycenterandscalefactork,yielding∆A'B'C'.∆A'B'C'iscongruentto∆DEFbySSScongruence.SincewecanmapΔABCtoΔDEFbyadilationandasequenceofisometries(thosethatmap∆A'B'C'to∆DEF,)thetrianglesaresimilarbydefinition.

b. SASSimilarity:Ifapairofsidesinonetriangleisproportionaltoapairofsidesinanothertriangle,andiftheanglesbetweenthosesidesarecongruent,thenthetrianglesaresimilar.Proof:Given∆ABCand∆DEFsuchthat!"!" =

!"!"=kand∠C=∠F.Dilate∆ABC

withanycenterandscalefactork,yielding∆A'B'C'.∆A'B'C'iscongruentto∆DEFbySAScongruence.Therefore,ΔABCandΔDEFaresimilarbythesameargumentasabove.

c. AASimilarity:Iftwoanglesinonetriangleareequaltotwoanglesinanothertriangle,thetrianglesaresimilar.Proof:Given∆ABCand∆DEFsuchthat∠C=∠Fand∠B=∠E.Let!"!" =k,sothatEF=ka.Dilate∆ABCfromanypointwithscalefactork.Theresulting∆A'B'C'iscongruentto∆DEFbyASAcongruence.Therefore,ΔABCandΔDEFaresimilar.



d. HLSimilarityCriterionforRightTriangles:Ifthehypotenuseandonelegofonerighttriangleareproportionaltothehypotenuseandonelegofanother,thentherighttrianglesaresimilar.Proof:Givenright∆ABCandright∆DEFsuchthat∠C=∠F=90˚,and!"!" =

!"!"=k.Dilate∆ABCfromanypointwithscalefactork.Theresulting

∆A'B'C'iscongruentto∆DEFbyHLcongruenceforrighttriangles.Therefore,ΔABCandΔDEFaresimilar.

13. Asegmentjoiningthemidpointsoftwosidesofatriangle(calledamidsegment)isparalleltothethirdsideandhalfaslong.Note:ThiswasprovedinChapter3.Weincludethisproofaswellbecauseitisshorterandmoreelegant.IfyouincludeTheorem14belowinyourdevelopment,youcanjustmentionthistheoremasacorollaryofit.Proof:LetDEbeamidsegmentofΔABC.DilateΔABCfromAwithscalefactor!!.SinceDandEaremidpoints,B'=DandC'=E.TheFTDtellsusthat!"||!"and!" = !

!!".14. Ifasegmentjoinspointsontwosidesofatrianglewhosedistancesarethesame

fractionk(0<k<1)ofthedistancefromtheircommonendpointtotheirotherendpoint,thenthesegmentjoiningthesepointsisparalleltothethirdsideanditslengthisthesamefractionkofit.Proof:LetDandEbepointsonsidesABandAEofΔABCrespectively,chosensothatAD=kABand!" = !"#,where0<k<1.DilateΔABCfromAwithscalefactor!.Bythedefinitionofdilation,B'=DandC'=E.TheFTDtellsusthatDE||BCandDE=kBC.Note:Asimilarargumentworksifk>1ifyoureplacesegmentsABandACbyraysABandAC.



GeneralTrapezoid15. PropertiesofaGeneralTrapezoid:

a. Apairofoppositesides(calledbasesinthedilationcase)areparallel.Proof:Ifonesideisadilationofanother,thecenterofdilationcan'tbeonalinecontainingoneofthesesides,becauseifitwere,theoppositesideswouldbecollinear.BytheFTD,theimageofasideisparalleltoitspre-image.Ifonesideisatranslationofanother,theimageofasegmentundertranslationbyavectornotparalleltothelineisalineparalleltoitspre-image.

b. Consecutiveangles(ondifferentbasesifthetrapezoidisnotaparallelogram)aresupplementary.Proof:Thisisapropertyofparallellinescutbyatransversal.



Chapter5:CirclesCircleDefinitions

Circle:Thesetofpointsthatareagivenfixeddistancefromagivenpoint(thecenter).Radius:Eitherthegivenfixeddistanceorasegmentjoiningthecentertoapointonacircle.Chord:Asegmentconnectingtwopointsonacircle.Diameter:Achordthatpassesthroughthecenterofacircle.Theworddiameteralsoreferstothelengthofanysuchsegment.Secant:Alinethatintersectsacircleintwopoints.Tangent:Alinethatintersectsacircleinexactlyonepoint(thepointoftangencyorpointofcontact).TangentSegment:Asegmentthatisasubsetofatangentline,withanendpointonthecircle.

Notation

Acirclecanbenamedbyitscenterifonlyonecircleinthediagramhasthatcenter.Inthatcase,CircleOnamesthecirclewhosecenterisO.

Theorems

Weincludeneithertheinscribedangletheoremnortheoremsaboutintersectingchords,becausewedidnotfindtransformationalproofsforthem.Theyshouldcertainlybeincludedinageometrycourse,buttheycanbeprovedusingtraditionalmethods.

1. Anydiameterofacircleisalineofsymmetry.

Proof:LetPbeapointoncircleOandletdbeadiameter.Considerreflectionind.ThereflectionofOPisOP',soOP'=OPbecausereflectionpreservesdistance.ThereforeP'isalsooncircleO.

2. Ifadiameterisperpendiculartoachord,itbisectsthechord.Proof:LetdiameterdofcircleObeperpendiculartochordAB.SinceAandBarebothonthecircle,OA=OB.Therefore,ΔOABisisosceles.ByTheorem1dofChapter2,itslineofsymmetryisaperpendicularbisector.

3. Ifadiameterbisectsachord,itisperpendiculartothechord.Proof:LetdiameterdofcircleObisectchordABatK.SinceAandBarebothonthecircle,OA=OB.Therefore,ΔOABisisosceles.ByTheorem1dofChapter2,itslineofsymmetryisaperpendicularbisector.

4. Theperpendicularbisectorofachordpassesthroughthecenterofthecircle.Proof:Thecenterofthecircleisequidistantfromtheendpointsofthechord,soitliesonitsperpendicularbisector.



5. Thereflectionofatangentsegmentinthesegmentjoiningitsexternalendpointtothecenterisanothertangentsegment.Proof:LetPAbeatangentsegmenttocircleOasshown.ReflectPAinPO;itsimageisPA'.Sincereflectionpreservesdistance,OA'=OA,soA'isoncircleO.NootherpointonthelinecontainingPAisadistancerfromO,sonootherpointonthelinecontainingPA'canberawayfromO.ThereforePA'isatangentsegment.

6. Tangentsegmentstoacirclefromanexternalpointareequal.Proof:UsingTheorem5andthesamediagram,PA'=PAbecausereflectionpreservesdistance.

7. Asegmentjoininganexternalpointtothecenterofacirclebisectstheangleformedbythetwotangentsegmentsdrawnfromthatpoint.Proof:UsingTheorem5andthesamediagram,∠OPA'=∠OPAbecausereflectionpreservesanglemeasure.

8. Alineperpendiculartoaradiusatitsendpointonthecircleisatangentline.Proof:LetlinetbeperpendiculartoradiusOPatP.treflectsontoitselfinOP.SupposetintersectedcircleOatanotherpointQ.ThenQ'wouldbeonbothtandcircleOaswell.Inthatcase,circleOandtwouldintersectinthreepoints(Q,PandQ'),whichcontradictsPostulate3.Therefore,tintersectscircleOonlyatP,soitisatangentline.

9. Atangenttoacircleisperpendiculartoaradiusdrawntothepointoftangency.Proof:LetubetangenttocircleO,withPasthepointoftangency.DropaperpendicularbfromOtou.LetP'bethereflectionofPinb.Sincebisperpendiculartou,P'mustbeonu.Butsincereflectionspreservedistance,OP=OP'.ThereforeP'isalsoonthecircle.Butuistangenttothecircleatonepoint,soP'=P.SincePisitsownreflectioninb,itmustbeonlineb.ThusOP,theradiusdrawntothepointoftangency,isperpendiculartou.

10. Whentwocirclesintersectintwopoints,thelinethroughtheircentersistheperpendicularbisectoroftheircommonchord.Proof:CirclesOandPhavecommonchordQR.SinceOandParebothequidistantfromQandR,theylieonitsperpendicularbisector.



Chapter6:PythagoreanTheoremTherearemanyproofsofthePythagoreantheorem,includingmanydissectionproofs.Thisoneistailor-madefortransformations,becauseitreliesonshowingthatquadrilaterals(nottriangles)arecongruent.Traditionalcurriculumdoesn’tincludecongruentquadrilaterals.Adissectionproofisveryvisual:youliterallycutthesquarebuiltonthelongerlegofarighttriangleintocongruentquadrilateralsandreassemblethepieces,alongwiththesquarebuiltontheshorterleg,tobuildasquareonthehypotenuse.Wewanttoprovethatthisreallyworks–thatthefivepiecesreallydoformasquareonthehypotenuse.Alongtheway,weusepropertiesofaparallelogramandwritesimpleequationsrelatingsomeofthesidelengths.HenryPerigal,Jr.(1April1801–6June1898)wasaBritishstockbrokerandamateurmathematician.HeprovidedthisdissectioninhisbookletGeometricDissectionsandTranspositions(London:Bell&Sons,1891).Hehadthedissectionprintedonhisbusinesscards,anditalsoappearsonhistombstone.(Source:Wikipedia)Thediagramshowsarighttrianglewithsquaresbuiltonthelegsandhypotenuse.OisthecenterofthesquareACED(theintersectionofitsdiagonals).DrawlinesthroughOparallelandperpendiculartoBAandcutthemoffatpointsP,Q,R,andSonthesquare.TherearefourrightangleswithvertexO.SquareADEChas4-foldrotationalsymmetry(Chapter2Theorem9d).Therefore,undera90˚counterclockwiserotationaroundO,A→D→E→C→A.ThisimpliesthatsidesAD→DE→EC→CA→AD.Inaddition,linePR→lineQS→linePR.SincePistheintersectionoflinePRandsideAD,andsinceQistheintersectionoflineQSandsideDE,thismeansthatP→Q.Asimilarargumentappliestotheotherpointsonthesidesofthesquare,soP→Q→R→S→P.SincetherotationisaroundO,Oremainsfixed.Sinceverticesmaptoverticesandrotationmapssegmentstosegments,wehaveshownthat,undera90˚counterclockwiserotationaroundO,OPDQ→OQER→ORCS→OSAP→OPDQ.Bydefinition,allfourofthesequadrilateralsarecongruent.Notethateachquadrilateralhastwooppositerightangles.



Translateeachofthesequadrilateralstoacornerofthesquarebuiltonthehypotenuse,asshowninthediagram.Byconstruction,RP||BA.AD||BEbecausetheoppositesidesofasquareareparallel.ByChapter2Theorems9aand7eandChapter3Theorem3b,BAPRisaparallelogram.ItsoppositesidesareequalbyChapter2Theorem4f.Therefore,a+d=e.Buttranslationpreservessegmentlength,soA'P'=eandC'P'=d.ThusA'C'=A'P'–C'P'=e–d=a.Thesameargumentappliestotheothersidesofthecentralwhitequadrilateral,soallsideshavelengthaandallanglesarerightangles,soitisasquarewithareaa2.(Itispossible,butnotnecessaryforthisproof,toshowitisthetranslationimageoftheoriginalsquareonsidea.)Bytherotationalsymmetry,OP=OQ=OR=OS=g,andsincetheoppositesidesofaparallelogramareequal,2g=c.Sinceeachquadrilateralhastworightanglesandthesumoftheanglesinaquadrilateralis360˚,theothertwoanglesaresupplementary.Sincethefourquadrilateralsarecongruent,∠BP'C'and∠AP'C'aresupplementary,soBP'andP'Aarecollinear.ThesameargumentappliestotheothermidpointsofsidesofthesquareonhypotenuseAB.Therefore,thefivepiecescoverthesquareonthehypotenusewithnogapsoroverlaps.Itfollowsthatthediagramshowsthattheareaofthesquarebuiltonthehypotenuseisthesumoftheareasofthesquaresbuiltonthelegs.FurtherExploration:Itispossibletofindthelengthseandfintermsofaandb.Thisisaninterestingexercise.

March 2018 © Lew Douglas and Henri Picciotto

www.MathEducationPage.org Appendix, p. 1

Appendix

An asterisk indicates a theorem which, in our view, should be discussed, but need not be proved formally in a standard geometry course. (See Chapter 1 for an explanation.)

List of Theorems from Chapter 1

Basic Theorems 1. *IfA'=Bunderareflection,thenB'=A.2. *Reflectionpreservescollinearityandbetweenness.3. *IfA→A'andB→B'underareflection,segmentABmustmapontosegmentA'B'.4. *Reflectionsmapraysontoraysandlinesontolines.5. *Congruentsegmentshaveequallength.Congruentangleshaveequalmeasure.6. *Thecorrespondingsidesandanglesofcongruentpolygonshaveequalmeasure.7. *ThereisareflectionthatmapsanygivenpointPontoanygivenpointQ.

Triangle Congruence 8. ApointPisequidistantfromtwopointsAandBifandonlyifitliesontheirperpendicular

bisector.9. IftwosegmentsABandCDhaveequallength,thenoneistheimageoftheother,withC

theimageofAandDtheimageofB,undereitheroneortworeflections.10. Equallengthsegmentsarecongruent.IfwecombinethiswithTheorem5,wehave:

Segmentsarecongruentifandonlyiftheyhaveequallength.11. CongruenceCriteriaforTriangles

a. SSSCongruence:Ifallsidesofonetrianglearecongruent,respectively,toallsidesofanother,thenthetrianglesarecongruent.

b. SASCongruence:Iftwosidesofonetrianglearecongruenttotwosidesofanother,andiftheincludedangleshaveequalmeasure,thenthetrianglesarecongruent.

c. ASACongruence:Iftwoanglesofonetrianglearecongruenttotwoanglesofanother,andifthesidescommontotheseanglesineachtrianglearecongruent,thenthetrianglesarecongruent.

12. HLCongruenceCriterionforRightTriangles:Ifthehypotenuseandonelegofonerighttrianglearecongruenttothehypotenuseandonelegofanother,thentherighttrianglesarecongruent.

13. Iftwotrianglesarecongruent,onecanbesuperimposedontheotherbyasequenceofatmostthreereflections.

14. *Angleswithequalmeasurearecongruent.IfwecombinethiswithTheorem5,wehave:Anglesarecongruentifandonlyiftheyhaveequalmeasure.

Two Reflections 15. *Ifalineisperpendiculartooneoftwoparallellines,itisperpendiculartotheother.16. Thecompositionoftworeflectionsinparallellinesistranslation.Thetranslationvectoris

perpendiculartothelines,pointsfromthefirstlinetothesecond,andhaslengthtwicethedistancebetweenthelines.Thisimpliesthatanytranslationcanbedecomposedintotworeflections.



17. Thecompositionoftworeflectionsinintersectinglinesisarotationaroundtheirpointofintersection.Theangleofrotationistwicethedirectedanglebetweenthelinesgoingfromthefirstreflectionlinetothesecond(eitherclockwiseorcounterclockwise).Thisimpliesthatanyrotationcanbedecomposedintotworeflections.

18. *Reflection,rotation,andtranslationpreservecollinearity,betweenness,segmentlengthandanglemeasure.

19. *IfA→A'andB→B'underareflection,rotation,ortranslation,segmentABmustmapontosegmentA'B'.

20. *Reflections,rotations,andtranslationsmapraysontoraysandlinesontolines.21. *Giventwointersectinglines,therearetworeflectionsmappingonetotheother.

Note:TherelatedtheoremfordistinctparallellinesisTheorem10ofChapter3.

Translations, Half-Turns, and Parallels 22. *IfA'=Bunderahalf-turn,thenB'=A.23. *TheimageofsegmentABunderahalf-turnarounditsmidpointisthesegmentBA.That

is,A'=BandB'=A.24. *Alineisitsownimageunderahalf-turnaroundapointontheline.25. Theimageofalineunderahalf-turnisparalleltothepre-image.26. *Whentwolinesintersect,theverticalanglesareequal.27. *Iftwodistinctlinesarecutbyatransversal,theyareparallelifandonlyifthealternate

interioranglesareequal.28. *Iftwodistinctlinesarecutbyatransversal,theyareparallelifandonlyifthe

correspondinganglesareequal.29. *Thecompositionoftranslationsiscommutative.30. *Thetranslationimageofalineisparalleltotheline.31. *Anyrepresentativeofavectorcanbesuperimposedonanyotherbyatranslation.32. *Iftwodistinctlinesarecutbyatransversal,anangleononelineisthetranslationimage

ofanangleontheotherifandonlyifthelinesareparallel.

Sum of Angles 33. Thesumoftheanglesofatriangleis180˚.Anexteriorangleofatriangleisequaltothe

sumoftheremoteinteriorangles.Therefore,itisgreaterthaneitherone.34. Thesumoftheinterioranglesofaquadrilateralis360˚.(Aconcavequadrilateralwill

haveaninterioranglegreaterthan180˚.)



List of Theorems from Chapter 4

Basic Theorems 1. *Adilationwithscalefactor-1isahalf-turnaroundthecenterofdilation.2. *Adilationwithscalefactork<0isthecompositionofadilationwithscalefactor-1and

adilationwithscalefactor|k|,allwiththesamecenter.3. *IfOisthecenterofadilation,kisthescalefactor,andO,A,andBaredistinctcollinear

points,thenA'B'=|k|AB.4. *Theimageofalineunderadilationisaline.5. FundamentalTheoremofDilations(FTD):IfC,A,andBarenotcollinear,thesegmentA'B'

joiningtheimagesofAandBunderadilationwithcenterCandscalefactorkisparalleltosegmentABandhaslength|k|AB.

6. Underadilation,A'B'=|k|AB.ThedirectedsegmentsABandA'B'pointinthesamedirectionifk>0andinoppositedirectionsifk<0.

7. *Dilationpreservesbetweenness.8. *Theimageofasegmentunderadilationisasegment,theimageofarayisaray.9. Dilationpreservesanglemeasure.10. Dilationpreservestheratioofthelengthsofanytwosegments.

Similar Triangles 11. Similartriangleshavecongruentanglesandproportionalsides.12. SimilarityCriteriaforTriangles

a. SSSSimilarity:Ifthesidesoftwotrianglesareproportional,thenthetrianglesaresimilar.

b. SASSimilarity:Ifapairofsidesinonetriangleisproportionaltoapairofsidesinanothertriangle,andiftheanglesbetweenthosesidesarecongruent,thenthetrianglesaresimilar.

c. AASimilarity:Iftwoanglesinonetriangleareequaltotwoanglesinanothertriangle,thetrianglesaresimilar.Note:Thesecorrespondtothecongruencecriteria.AASimilaritycorrespondstoASACongruence.Thesideisn’tneededbecauseittakestwoormoresidestomakeaproportion.

13. Asegmentjoiningthemidpointsoftwosidesofatriangle(calledamidsegment)isparalleltothethirdsideandhalfaslong.

14. Ifasegmentjoinspointsontwosidesofatrianglewhosedistancesarethesamefractionk(0<k<1)ofthedistancefromtheircommonendpointtotheirotherendpoint,thenthesegmentjoiningthesepointsisparalleltothethirdsideanditslengthisthesamefractionkofit.

General Trapezoid 15. PropertiesofaGeneralTrapezoid:

a. Apairofoppositesides(calledbasesinthedilationcase)areparallel.b. Consecutiveangles(ondifferentbasesifthisisnotaparallelogram)are

supplementary.



List of Theorems from Chapter 5 1. Anydiameterofacircleisalineofsymmetry.2. Ifadiameterisperpendiculartoachord,itbisectsthechord.3. Ifadiameterbisectsachord,itisperpendiculartothechord.4. Theperpendicularbisectorofachordpassesthroughthecenterofthecircle.5. Thereflectionofatangentsegmentinthesegmentjoiningitsexternalendpointtothe

centerisanothertangentsegment.6. Tangentsegmentsterminatingonacirclefromanexternalpointareequal.7. Asegmentjoininganexternalpointtothecenterofacirclebisectstheangleformedby

thetwotangentsegmentsdrawnfromthatpoint.8. Alineperpendiculartoaradiusatitsendpointisatangentline.9. Alinenotperpendiculartoaradiusatitsendpointisnotatangentline.(Theorems8and

9canbecombined:Alineisperpendiculartoaradiusatitsendpointifandonlyifitisatangentline.)

10. Whentwocirclesintersectintwopoints,thelinethroughtheircentersistheperpendicularbisectoroftheircommonchord.

Transformational Proof in High School Geometry · 2018-03-30 · Preface p. 1 Preface:...

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