Proving Triangles and Quadrilaterals Satisfy … Triangles and Quadrilaterals Satisfy Transformational Definitions 1. Definition of Isosceles Triangle: A triangle with one line of

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ProvingTrianglesandQuadrilateralsSatisfyTransformationalDefinitions

1. DefinitionofIsoscelesTriangle:Atrianglewithonelineofsymmetry.

a. Ifatrianglehastwoequalsides,itis

isosceles.Proof:LetABandACbetheequalsides.AmustlieontheperpendicularbisectorlofBCbecauseitisequidistantfromBandC.Bythedefinitionofreflection,B'=Cunderreflectioninl.A'=Abecauseitliesonl.Therefore,lisalineofsymmetryforΔABC.

b. Ifatrianglehastwoequalangles,itisisosceles.Proof:DrawrayAD,theanglebisectorof∠BAC.Twoanglesof∆BADand∆CADareequal,sothethirdangles(∠BDAand∠CDA)mustbeequal.Sincetheyaresupplementary,theyarebothrightangles.ReflectBinrayAD.Sincereflectionspreserveangles,B'mustbeonrayAC.Bythedefinitionofreflection,B'mustbeonrayBD.Therefore,B'isattheintersectionofrayBDandrayAC,whichisC.Therefore,B'=C.SinceCisthereflectionofBinlineAD,andAisitsownreflectioninlineAD,ADisalineofsymmetryforthetriangle.

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.

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

2. DefinitionofEquilateralTriangle:Atrianglewithtwolinesofsymmetry.

a. Ifatrianglehasallsidesequal,thenit'sanequilateraltriangle.Proof:InΔABC,AB=BC=CA.SinceAB=AC,thetriangleisisosceleswithsymmetrylinem.SinceCA=CB,thetriangleisisosceleswithsymmetrylinen.Sinceithastwolinesofsymmetry,itisequilateral.

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

3. DefinitionofParallelogram:Aquadrilateralwith2-foldrotationalsymmetry.

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

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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.

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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. DefinitionofKite:Aquadrilateralwithonelineofsymmetrythroughoppositevertices.b. Iftwodisjointpairsofconsecutivesidesofa

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

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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,andreflectABCDinl.SinceAandClieonl,A'=AandC'=C.Bythedefinitionofreflection,B'=DandD'=B.Therefore,lisalineofsymmetryandABCDisakite.

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5. DefinitionofIsoscelesTrapezoid:Aquadrilateralwithalineofsymmetrythoughmidpointsofoppositesides.

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

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∠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'=CandC'=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.ThisperpendicularbisectoristhereforealineofsymmetryforABCD,soitisanisoscelestrapezoidbydefinition.

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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.ReflectBacrossdiagonalAC.Becauseoftheequalangles,B'liesonbothrayADandrayCD,soB'=D.Sincereflectionpreservesdistance,AD=AB.Buttheoppositesidesofa

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parallelogramareequal,soABCDisequilateral.Therefore,ABCDisarhombusbyTheorem6b.

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. DefinitionofRectangle: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.

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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||DC.Sincethediagonalsareequalaswell,ABCDisanisoscelestrapezoidwhoselineofsymmetrypassesthroughmidpointsofABandDCbyTheorem5e.BythesameargumentwithparallelsidesADandBC,ABCDisanisoscelestrapezoidwhoselineofsymmetrypassesthroughmidpointsofADandBC.ItfollowsthatABCDsatisfiesthesymmetrydefinitionofarectangle.

8. DefinitionofSquare:Aquadrilateralwithfourlinesofsymmetry:twopassingthroughoppositeverticesandtwopassingthroughmidpointsofoppositesides.

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

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

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c. Anequilateralquadrilateralwitharightangleisasquare.Proof:Anequilateralquadrilateralisarhombus.Oppositeanglesofaquadrilateralareequalandthesumoftheanglesis360˚,soallanglesarerightanglesandthequadrilateralisalsoequiangular.Anequiangularquadrilateralisarectangle.Ifaquadrilateralisbotharhombusandarectangle,ithasfourlinesofsymmetryandisthereforeasquare.

d. Anequiangularquadrilateralwithconsecutiveequalsidesisasquare.Proof:Anequiangularquadrilateralisarectangle.Theoppositesidesofarectangleareequal,andifconsecutivesidesarealsoequal,itmustbeequilateral.Anequilateralquadrilateralisarhombus.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'aswell.BecausethediagonalsofquadrilateralABA'Cbisecteachother,itisaparallelogrambyTheorem3a.Butaparallelogramwitharightangleisarectangle,andthediagonalsofarectangleareequal.ThusAD=½AA'=½BC.

b. Asegmentjoiningthemidpointsoftwosidesofatriangle(calledamidsegment)isparalleltothethirdsideandhalfaslong.Proof:IntriangleABC,DandEaremidpointsofACandBCrespectively.RotateΔABCandsegmentDE180˚aroundpointE.SinceEisamidpoint,B'=CandC'=B.Therefore,quadrilateralABA'Chas2-foldrotationalsymmetry,soitisaparallelogram.

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BecauserotationpreservessegmentlengthandDisamidpoint,AD=DC=D'B.ButADisparalleltoBD'aswell,soABD'DisalsoaparallelogrambyTheorem3d.Theoppositesidesofaparallelogramareparallel,soDE||AB.Becauserotationpreserveslength,DE=ED'=½DD',andbecausetheoppositesidesofaparallelogramareequal,DD'=AB.HenceDE=½AB.Note:Theproofisshorterandmoreelegantusingdilation.