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November2016 ©LewDouglasandHenriPicciotto ProvingTrianglesandQuadrilaterals SatisfyTransformationalDefinitions 1. DefinitionofIsoscelesTriangle:Atrianglewithonelineofsymmetry. a. Ifatrianglehastwoequalsides,itis isosceles. Proof:LetABandACbetheequalsides.A mustlieontheperpendicularbisectorlof BCbecauseitisequidistantfromBandC. Bythedefinitionofreflection,B'=Cunder reflectioninl.A'=Abecauseitliesonl. ThereforelisalineofsymmetryforΔABC. b. Ifatrianglehastwoequalangles,itis isosceles. Proof:DrawrayAD,theanglebisectorof ∠BAC.Twoanglesof∆BADand∆CADare equal,sothethirdangles(∠BDAand ∠CDA)mustbeequal.Sincetheyare supplementary,theyarebothrightangles. ReflectBinAD.Sincereflectionspreserve angles,B’mustbeonrayAC.Bythe definitionofreflection,B’mustbeonline BD.Therefore,B’isattheintersectionof BDandAC,whichisC.Therefore,B’=C. SinceCisthereflectionofBinAD,andAis itsownreflectioninAD,ADisalineofsymmetryforthetriangle. c. Ifananglebisectorofatriangleisalsoanaltitude,thetriangleis isosceles. Proof:Letl,thebisectorof∠AbeperpendiculartosideBCatD,sothat ∠DAB=∠DAC.ReflectBinlwithimageB'.BecauseBC⊥AD,B'liesonray BC.Because∠DAB=∠DACandreflectionpreservesangles,italsolieson rayAC.Sincetwolinescanonlyintersectatonepoint,B'=C.Thismeans thatlisalineofsymmetryforΔABC. d. Ifanaltitudeofatriangleisalsoamedian,the triangleisisosceles. Proof:SincealtitudeADisalsoamedian,ADis theperpendicularbisectorofBC.Sinceanypoint ontheperpendicularbisectorofasegmentis equidistantfromtheendpoints,AB=AC.By www.MathEducationPage.org p.1 November2016 ©LewDouglasandHenriPicciotto theorem(a)above,thetriangleisisosceles. e. Ifananglebisectorofatriangleisalsoamedian,thetriangleisisosceles. Proof:Postponeduntiltheendoftherhombussectionbecausearhombus isconstructedintheproof. 2. DefinitionofEquilateralTriangle:Atrianglewithtwolinesofsymmetry. a. Ifatrianglehasallsidesequal,thenit'san equilateraltriangle. Proof:InΔABC,AB=BC=CA.Since AB=AC,thetriangleisisosceleswith symmetrylinem.SinceCA=CB,the triangleisisosceleswithsymmetrylinen. Sinceithastwolinesofsymmetry,itis equilateral. b. Ifatrianglehasallanglesequal,thenit'sanequilateraltriangle. Proof:Theargumentisvirtuallyidenticaltothepreviousone,sincea trianglewithtwoequalanglesisisosceles. 3. DefinitionofParallelogram:Aquadrilateralwith2-foldrotationalsymmetry. a. Ifthediagonalsofaquadrilateralbisecteachother,thequadrilateralisa parallelogram. Proof:RotatequadrilateralABCD180˚ aroundpointE,theintersectionofthe diagonals.Sincetherotationisthrougha straightangle,B'liesonrayBDonthe oppositesideofEfromB.Sincerotation preservesdistance,B'=D.SimilarlyA'=C. Bythesameargument,C'=AandE'=B. Becauserotationmapssegmentsto segments,quadrilateralABCDhas2-foldrotationalsymmetry.Therefore, itisaparallelogram. www.MathEducationPage.org p.2 November2016 ©LewDouglasandHenriPicciotto b. Ifoppositesidesofaquadrilateralareparallel,thequadrilateralisa parallelogram. Proof:GivenquadrilateralABCDasinthisfigure.Wewouldliketoprove thatithas2-fold(half-turn)symmetry. LetMbethemidpointof AC : ConsiderH,thehalf-turnwithcenterM.SinceMisthemidpointof segmentAC,A'=CandC'=AunderH.BecauseMisonneitherknorl, theirimagelinesareparalleltotheirpre-images,andbecauseofthe parallelpostulate,thereisonlyoneparalleltokthroughCandone paralleltolthroughA.Therefore,k'=m,andl'=n.Bistheintersectionof lineskandl,andthereforeitsimageistheintersectionoflinesmandn. Therefore,DistheimageofB.SotheimageofquadrilateralABCDinthe half-turnHisitself. www.MathEducationPage.org p.3 November2016 ©LewDouglasandHenriPicciotto c. Ifoppositesidesofa quadrilateralareequal,the quadrilateralisa parallelogram. Proof:InquadrilateralABCD,drawdiagonalACanditsmidpointE.Under ahalfturnaroundE,A'=CandC'=A.SinceCB=AD,B'liesoncircleA withradiusAD.SinceAB=CD,B'liesoncircleCwithradiusCD.These circlesintersectatDandF.ButFisonthesamesideoflineACasB,soB' ≠F.ThereforeB'=DandABCDhas2-foldsymmetryaroundE.By definitionABCDisaparallelogram. d. Iftwosidesofaquadrilateralareequalandparallel,thequadrilateralisa parallelogram. Proof:Inquadrilateral ABCD,supposeABisequal andparalleltoDC.Draw diagonalBDandits midpointM.UnderahalfturnaroundM,B'=Dand D'=B.TheimageofrayBA isparalleltoAB,soitmustcoincidewithrayDC.BecauseAB=DC,that meansthatA'=C,andthereforeC'=A.HenceABCDisaparallelogram. 4. DefinitionofKite:Aquadrilateralwithonelineof symmetrythroughoppositevertices. a. Iftwodisjointpairsofconsecutivesidesofa quadrilateralareequal,thequadrilateralisakite. Proof:InquadrilateralKITE,KI=KEand TI=TE.SinceKandTarebothequidistantfromI andE,theylieintheperpendicularbisectorof segmentIE.Therefore,listheperpendicular bisectorofthissegment.Underreflectioninl,I'=E andE'=I.BecauseKandTbothlieonl,K'=Kand T'=T.ListhereforealineofsymmetryandKITEis akite. www.MathEducationPage.org p.4 November2016 ©LewDouglasandHenriPicciotto b. Ifadiagonalofaquadrilateralbisectsapair ofoppositeangles,thequadrilateralisakite. Proof:LabelaslthediagonalKTthatbisects∠IKE and∠ITE.Considerreflectioninl.SinceKandTare onl,K'=KandT'=T.Since∠IKT=∠EKT,I'lieson rayKE.Since∠ITK=∠ETK,I'liesonrayTE.Because theseraysintersectatE,I'=E.Similarly, E'=I.Therefore,lisalineofsymmetryandKITEisa kite. c. Ifonediagonalofaquadrilateralperpendicularly bisectstheother,thequadrilateralisakite. Proof:InquadrilateralKITE,KTperpendicularly bisectsIE.LetlbethelinethroughKandT,and reflectKITEinl.SinceKandTlieonl,K'=kandT' =T.Bythedefinitionofreflection,I'=EandE'=I. Therefore,lisalineofsymmetryandKITEisakite. 5. DefinitionofIsoscelesTrapezoid:Aquadrilateralwith alineofsymmetrythoughmidpointsofoppositesides. a. Ifonepairofoppositesidesofaquadrilateralareparallelandapairof consecutiveanglesononeofthesesidesareequal,thequadrilateralisan isoscelestrapezoid. Proof:Inquadrilateral TRAP,PA||TRand∠T=∠R. Letlbetheperpendicular bisectorofTR.Under reflectioninl,T'=Rand R'=T.SincePA||TR,PA⊥l. Therefore,P'liesonrayPA. BecauseT'=R,TRmapsto RT,∠T=∠R,andreflection preservesangles,P'lieson rayRA.Theseraysintersect atA,soP'=A.ThuslisalineofsymmetryforTRAP,soTRAPisan isoscelestrapezoid. www.MathEducationPage.org p.5 November2016 ©LewDouglasandHenriPicciotto b. Iftwodisjointpairsof consecutiveanglesofa quadrilateralareequal,the quadrilateralisanisosceles trapezoid. Proof:Inquadrilateral TRAP,∠T=∠Rand∠P= ∠A.Thisproofdependson thefactthatthesumofthe interioranglesofa quadrilateralis360˚.Given thatandtwopairsofequal angles,itfollowsthat∠T+ ∠P=180˚.Asaconsequence,PA||TR.Nowwehaveprovedthe hypothesesofpreviousTheorem5a,soTRAPisanisoscelestrapezoid. c. Iftwooppositesides ofaquadrilateralare parallelandifthe othertwosidesare equalbutnot parallel,thenthe quadrilateralisan isoscelestrapezoid. Proof:In quadrilateralTRAP,PA||TR,TP=RA,andTPisnotparalleltoRA. ThroughR,drawalineparalleltoPTmeetingrayPAatB.SinceTRBPhas twopairsofoppositeparallelsides,itisaparallelogram.Becausethe oppositeanglesofaparallelogramareequal,∠T=∠RBA.Theopposite sidesofaparallelogramarealsoequal,soTP=RB=RA.Iftwosidesofa triangleareequal,thetriangleisisosceles,whichimpliesthat∠RBA= ∠RAB.Finally,becausePA||TR,∠RAB=∠TRA.Thechainofequalangles nowreads∠T=∠RBA=∠RAB=∠TRA.ThismeansthatTRAPhasapair ofconsecutiveequalanglesononeofitsparallelsides.ByTheorem5a, TRAPisanisoscelestrapezoid. d. Ifalineperpendicularlybisectstwosidesofaquadrilateral,the quadrilateralisanisoscelestrapezoid. Proof:Thetwosidesreferredtocan'tbeconsecutive,becauseifthey were,youwouldhavetwoconsecutiveparallelsides,whichcouldn't intersect.InquadrilateralTRAP,listheperpendicularbisectorofTRand PA.Underreflectioninl,therefore,P'=Aand T'=B.Thuslisalineofsymmetry,whichisthedefinitionofanisosceles trapezoid. www.MathEducationPage.org p.6 November2016 ©LewDouglasandHenriPicciotto e. Iftwosidesofaquadrilateralareparallelandifthediagonalsareequal, thequadrilateralisanisoscelestrapezoid. Proof: InquadrilateralTRAP,PA||TRandTA=RP.LetMbethemidpointofRA. RotateTRAP180˚aroundM.SinceA'=RandR'=A,theimage quadrilateralisT'ARP'.Sinceahalfturnaroundapointnotonaline producesanimageparalleltothepre-image,sincePA||TR,andsince A'=RandR'=A;T,RandP'arecollinear,asareP,AandT'.Because rotationpreservessegmentlength,RP=AP'.Therefore,TA=RP=AP'. ConsiderΔTAP'.Sincetwosidesareequal,itisisosceles,so ∠RTA=∠RP'A.RPisalsorotated180˚aroundapointnotontheline,so imageisparalleltopre-image.Thatis,PR||P'A.UsingtransversalTP',we seethat∠RP'A=∠TRP.Thus∠RTA=∠RP'A=∠TRP.Becausetwoangles inΔTRBareequal,thetriangleisisosceles,whichimpliesthatBT=BR.In otherwords,TisequidistantfromTandR.Becauseofthis,itmustlieon theperpendicularbisectorofTR.AsimilarargumentshowsthatBlieson theperpendicularbisectorofPA.SincetheperpendicularbisectorsofTR andPApassthroughthesamepointB,theycoincide.Thisperpendicular bisectoristhereforealineofsymmetryforTRAP,whichisanisosceles trapezoidbydefinition. 6. Rhombus:Aquadrilateralwithtwolinesofsymmetrypassingthroughopposite vertices.(Soarhombusisakiteintwodifferentways.) a. Ifthediagonalsofaquadrilateralperpendicularlybisecteachother,the quadrilateralisarhombus. Proof:Bythedefinitionofreflection,thetwoverticesnotoneither diagonalareimagesofeachotherunderreflectioninthatdiagonal. Therefore,bothdiagonalsarelinesofsymmetry,whichisthedefinitionof arhombus. b. Ifaquadrilateralisequilateral,itisarhombus. Proof:Sinceoppositesidesareequal,thequadrilateralisaparallelogram. Therefore,thediagonalsbisecteachother.Sincetwodisjointpairsof consecutivesidesareequal,thequadrilateralisakite.Therefore,the diagonalsareperpendicular.Nowweknowthatthediagonals perpendicularlybisecteachother,sothequadrilateralisarhombusby Theorem6a. www.MathEducationPage.org p.7 November2016 ©LewDouglasandHenriPicciotto c. Ifbothdiagonalsofaquadrilateralbisectapairofoppositeangles,the quadrilateralisarhombus. Proof:Considerthediagonalsseparately. Since∠Band∠Darebisected,ABCDisa kitewithAD=CDandAB=CB.Since∠A and∠Carebisected,ABCDisakitewith AD=ABandCD=CB.Therefore,allfour sidesareequalandthequadrilateralisa rhombusbyTheorem6b. d. Ifbothpairsofoppositesidesofaquadrilateralareparallelandiftwo consecutivesidesareequal,thequadrilateralisarhombus. Proof:Sincebothpairsofoppositesidesareparallel,thequadrilateralisa parallelogram.Therefore,bothpairsofoppositesidesareequal.Sincea pairofconsecutivesidesareequal,allfoursidesmustbeequal.Hencethe quadrilateralisarhombusbyTheorem6b. e. Ifadiagonalofaparallelogrambisectsanangle,theparallelogramisa rhombus. Proof:InparallelogramABCD, diagonalACbisects∠DAB.The oppositesidesofaparallelogramare parallel,sothisimpliesthat∠DCBis bisectedaswellbyanglepropertiesof parallellines.ReflectBacrossAC. Becauseoftheequalangles,B'lieson bothrayADandrayCD,soB'=D.Since reflectionpreservesdistance,AD=AB.Buttheoppositesidesofa parallelogramareequal,soABCDisequilateral.Therefore,ABCDisa rhombus. f. Ifananglebisectorofatriangleisalsoa median,thetriangleisisosceles. Proof:ΔABC,rayADbisects∠BACand BD=CD.RotateΔABC180˚aroundD.Because DisthemidpointofBCandtherotationisa half-turn,B’=DandD’=B.Becauserotation preservesdistance,AD=A'D.Nowthe diagonalsofquadrilateralABA'Cbisecteach other,soABA'Cisaparallelogram.But diagonalAA'bisects∠BAC,sobyTheorem6e, ABA'Cisarhombus.Arhombusisequilateral, soAB=AC. www.MathEducationPage.org p.8 November2016 ©LewDouglasandHenriPicciotto 7. DefinitionofRectangle:Aquadrilateralwithtwolinesofsymmetrypassing throughmidpointsoftheoppositesides.(Soarectangleisanisoscelestrapezoid intwodifferentways.) a. Ifaquadrilateralisequiangular,itisarectangle. Proof:Because∠A=∠Band∠C=∠D,ABCDisanisoscelestrapezoidwith lineofsymmetrythrough midpointsofABandDC. Because∠A=∠Dand∠B=∠C, ABCDisanisoscelestrapezoid withlineofsymmetrythrough midpointsofADandBC.By definition,ABCDisarectangle. b. Ifaparallelogramhasaright angle,thentheparallelogramis arectangle. Proof:Suppose∠A=90˚.Then∠C=90˚becauseoppositeanglesofa parallelogramareequal.Thesumoftheinterioranglesofaquadrilateral is360˚,whichleavesatotalof180˚for∠Band∠D.Sincetheyarealso equal,theymustberightanglesalso.Henceallanglesareequalright anglesandthequadrilateralisarectanglebyTheorem7a. c. Anisoscelestrapezoidwitharightangleisarectangle. Proof:SupposeABCDisanisosceles trapezoidwithlineofsymmetry passingthroughbasesABandDC. Withoutlossofgenerality,wecan supposethat∠A=90˚.Becausethe basesofanisoscelestrapezoidare paralleland∠EDCand∠EABare translationimagesofeachother, ∠EDC=∠EAB=90˚.∠ADCand∠EDC aresupplementary,so∠ADC=90˚also. Wealsoknowthattwoconsecutive anglesofanisoscelestrapezoidonthe samebaseareequal,so∠B=∠A=90˚and∠C=∠ADC=90˚.NowABCDis equiangular,and,byTheorem7a,anequiangularquadrilateralisa rectangle. www.MathEducationPage.org p.9 November2016 ©LewDouglasandHenriPicciotto d. Ifthediagonalsofaparallelogramare equal,theparallelogramisarectangle. Proof:BecauseABCDisaparallelogram,AB||DC.Sincethediagonalsare equalaswell,ABCDisanisoscelestrapezoidwhoselineofsymmetry passesthroughmidpointsofABandDC.Bythesameargumentwith parallelsidesADandBC,ABCDisanisoscelestrapezoidwhoselineof symmetrypassesthroughmidpointsofADandBC.ItfollowsthatABCD satisfiesthesymmetrydefinitionofarectangle. 8. DefinitionofSquare:Aquadrilateralwithfourlinesofsymmetry:twopassing throughoppositeverticesandtwopassingthroughmidpointsofoppositesides. a. Arectanglewithconsecutiveequalsidesisasquare. Proof:Arectanglehastwolinesofsymmetrypassingthroughthe oppositesides.Theoppositesidesofarectangleareequal,soiftwo consecutivesidesarealsoequal,itisequilateral.Anequilateral quadrilateralisarhombus,soitsdiagonalsareadditionallinesof symmetry.Therefore,therectangleisasquare. b. Arhombuswithconsecutiveequalanglesisasquare. Proof:Thediagonalsofarhombusarelinesofsymmetry.Theopposite anglesofarhombusareequal,soiftwoconsecutiveanglesarealsoequal, itisequiangular.Anequiangularquadrilateralisarectangle,soithastwo additionallinesofsymmetrypassingthroughoppositesides.Therefore, therhombusisasquare c. Anequilateralquadrilateralwitharightangleisasquare. Proof:Anequilateralquadrilateralisarhombus.Oppositeanglesofa quadrilateralareequalandthesumoftheanglesis360˚.Fromthis,it followsthatallanglesarerightanglesandthequadrilateralisalso equiangular.Anequiangularquadrilateralisarectangle.Ifaquadrilateral isbotharhombusandarectangle,ithasfourlinesofsymmetryandis thereforeasquare. d. Anequiangularquadrilateralwithconsecutiveequalsidesisasquare. Proof:Anequiangularquadrilateralisarectangle.Theoppositesidesofa rectangleareequal,andifconsecutivesidesarealsoequal,itmustbe equilateral.Anequilateralquadrilateralisarhombus.Ifaquadrilateralis botharhombusandarectangle,ithasfourlinesofsymmetryandis thereforeasquare. www.MathEducationPage.org p.10 November2016 ©LewDouglasandHenriPicciotto e. Aquadrilateralwith4-foldrotationalsymmetryisasquare. Proof:Sinceaquadrilateralhasfoursides,consecutivesidesandangles mustmaptoeachotherundera90˚rotation.Becauserotationspreserve sidesandangles,thequadrilateralmustbebothequilateraland equiangular,whichimpliesthatitisbotharectangleandarhombus.Ifa quadrilateralisbotharhombusandarectangle,ithasfourlinesof symmetryandisthereforeasquare. 9. Puttingitalltogether: a. ARightTriangleProperty Themediantothehypotenuseofaright trianglehashalfthelengthofthehypotenuse. Proof:InrighttriangleABC,BD=CD.Rotate ΔABCandmedianAD180˚aroundD.Because rotationspreservedistanceandtherotationis 180˚,DisthemidpointofAA'aswellasBC.Becausethediagonalsof quadrilateralABA'Cbisecteachother,itisaparallelogram.Buta parallelogramwitharightangleisarectangle,andthediagonalsofa rectangleareequal.ThusAD=½AA'=½BC. b. MidsegmentTheoremforaTriangle Asegmentjoiningthemidpointsoftwosidesofatriangleisparallelto thethirdsideandhalfaslong. Proof:Intriangle ABC,DandEare midpointsofACand BCrespectively. RotateΔABCand segmentDE180˚ aroundpointE.Since Eisamidpoint,B'=C andC'=B.Therefore, quadrilateralABA'C has2-foldrotationalsymmetry,soitisaparallelogram.Becauserotation preservessegmentlengthandDisamidpoint,AD=DC=D'B.ButADis paralleltoBD'aswell,soABD'DisalsoaparallelogrambyTheorem3d. Theoppositesidesofaparallelogramareparallel,soDE||AB.Because rotationpreserveslength,DE=ED'=½DD',andbecausetheopposite sidesofaparallelogramareequal,DD'=AB.HenceDE=½AB. Note:Theproofwouldbeshorterandmoreelegantusingdilationrather thanisometries. www.MathEducationPage.org p.11