Download Proving Triangles and Quadrilaterals are Special

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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
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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