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.
e. Ifananglebisectorofatriangleisalsoamedian,thetriangleisisosceles.Proof:Postponeduntiltheendoftherhombussection(afterTheorem6e),becausearhombusisconstructedintheproof.
a. Ifatrianglehasallsidesequal,thenit'sanequilateraltriangle.Proof:InΔABC,AB=BC=CA.SinceAB=AC,thetriangleisisosceleswithsymmetrylinem.SinceCA=CB,thetriangleisisosceleswithsymmetrylinen.Sinceithastwolinesofsymmetry,itisequilateral.
b. Ifatrianglehasallanglesequal,thenit'sanequilateraltriangle.Proof:Theargumentisvirtuallyidenticaltothepreviousone,butusesTheorem1binsteadof1a.
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.
b. Ifoppositesidesofaquadrilateralareparallel,thequadrilateralisaparallelogram.Proof:GivenquadrilateralABCDasinthisfigure,withsidesextendedtolinesk,l,m,andn.Wewouldliketoprovethatithas2-fold(half-turn)symmetry.
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.
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.
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
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.
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
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||DC.Sincethediagonalsareequalaswell,ABCDisanisoscelestrapezoidwhoselineofsymmetrypassesthroughmidpointsofABandDCbyTheorem5e.BythesameargumentwithparallelsidesADandBC,ABCDisanisoscelestrapezoidwhoselineofsymmetrypassesthroughmidpointsofADandBC.ItfollowsthatABCDsatisfiesthesymmetrydefinitionofarectangle.
a. Arectanglewithconsecutiveequalsidesisasquare.Proof:Arectanglehastwolinesofsymmetrypassingthroughtheoppositesides.Theoppositesidesofarectangleareequal,soiftwoconsecutivesidesarealsoequal,itisequilateral.Anequilateralquadrilateralisarhombus,soitsdiagonalsareadditionallinesofsymmetry.Therefore,therectangleisasquare.
b. Arhombuswithconsecutiveequalanglesisasquare.Proof:Thediagonalsofarhombusarelinesofsymmetry.Theoppositeanglesofarhombusareequal,soiftwoconsecutiveanglesarealsoequal,itisequiangular.Anequiangularquadrilateralisarectangle,soithastwoadditionallinesofsymmetrypassingthroughoppositesides.Therefore,therhombusisasquare
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.
b. Asegmentjoiningthemidpointsoftwosidesofatriangle(calledamidsegment)isparalleltothethirdsideandhalfaslong.Proof:IntriangleABC,DandEaremidpointsofACandBCrespectively.RotateΔABCandsegmentDE180˚aroundpointE.SinceEisamidpoint,B'=CandC'=B.Therefore,quadrilateralABA'Chas2-foldrotationalsymmetry,soitisaparallelogram.
