MAT211:FinalExamReviewStudent‐WrittenQuestionsFall2010
Chapter1(Intro)1. Atriangleisisoscelesifandonlyifthebaseanglesarecongruent.(KatySpencer)2. Provethatthediagonalsofaparallelogrambisectoneanother(reliesonEuclid’s
5thPostulate)(Maggie)3. Aquadrilateralissaidtobecyclicifthereisacirclepassingthroughallitsfour
vertices.Givenaquadrilateraliscyclic,provethatthesumofthemeasuresofitsnon‐adjacentinterioranglesis180degrees.(Greg)
Finitegeometry(Fe‐Fo)4. Giventhefollowingsetofaxioms,writethreeconjecturesandproveoneofthem.
(RyanD.)Axiom1:Thereexistexactlynpointsinthesystem.Axiom2:Everysetoftwopointsliesonexactlyoneline.Axiom3:Eachlinecontainsexactlytwopoints.
5. Provetheconjecturefromthegivenaxioms.(StephanieM.)A1:Thereisagroupoffiveathletes.A2:Foreverythreeathletes,thereisacommonsporttheyhaveshared.A3:Notalloftheathleteshaveplayedthesamesport.Conjecture:Thereare10possiblesportsthathavebeenshared.
6. TherearemanydifferenttheoremsthatcanbeprovedfromtheAxiomsbelow.Proveoneofthetheoremsprovided.(StephanieS.)Axiom1.Thereexistexactlythreedistinctstudents,andthreedistinctcolleges.Axiom2.Twodistinctstudentsbelongtoexactlyonecollege.Axiom3.Notallstudentsbelongtothesamecollege.Axiom4.Anytwodistinctcollegescontainatleastonestudentwhichbelongstoboth.Theorem1.Twodistinctcollegescontainexactlyonestudent.Theorem2.Acollegecannotcontainthreedistinctstudents.Theorem3.Thereexistsasetoftwocollegesthatcontainallthestudents.
Chapter2(Triangles)7. ConsiderthetriangleABCwithincenterI.Provethattheincenterisapointthat
isequidistantfromeachsideofthetriangle.(Ellen)8. Determinewhetherthefollowingstatementsaretrueorfalse.Ifthestatementis
animplicationstatethehypothesisandconclusion.Thenstatetheconverseandthecontrapositive.(Toni)1.IfAisaparallelogramthanthediagonalsbisecteachother.2.IfAisarighttrianglethanitisalwaysanisoscelestriangle3.Thecenterofacircleliesontheperpendicularbisectoroftwopoints4.Allrhombiaresquares5.Everyrectanglehasthreesides,andallrighttrianglesareequiangular.
9. AssumingthatSASistrue,proveASA.(Anna)
10. ProvethatanyXontheperpendicularbisectorofsegmentABisthecenterofacirclethroughpointsAandB.(Ketty)
Chapter3(Circles)11. Twocirclesaresaidtobeorthogonaliftheirtangentsareperpendicularattheir
pointofintersection.(Clarissa)a.Describehowtoconstructorthogonalcircles.b.Provethatthereareperpendiculartangentsattheotherpointofintersectionofthecircles.
12. ProveorDisprovethattheareaofthesemi‐circlesonthelegsofarighttriangle(AB&BC)sumtotheareaofthesemi‐circleonthehypotenuse(CA).(Mike)
13. GiventriangleXYZ,extendsidesXYandXZcreatingexterioranglesatYandZ.ProvethatthebisectorsoftheexterioranglesatYandZareconcurrentwiththebisectoroftheinteriorangleatX.CallthispointofintersectionR.(Caitlin)
14. ProvethattheoppositeanglesinacyclicquadrilateralABCDaresupplementary.(JoeK.)
Chapter4(Analyticgeometry)15. Prove the Pythagorean theorem using analytic geometry. Be sure to justify that the
points youusemakearighttriangle.(Adam)16. Provethemidpointformula(Kevin)17. Provethatthediagonalsofarhombusareperpendicularbisectorsofeachother
(KatieSkerik)18. (Jamie)
a.)Usingcoordinates,constructarighttriangle.b.)Findthemidpointofthehypotenuse.c.)Prove(usingthecoordinates)thatthemidpointofthehypotenuseisequidistancefromalltheverticesofthetriangle.
Chapter6(Transformationalgeometry)19. Listtheorientationandfixedpointsifanyforthefollowingcompositionof
isometries.(Justin)1. Reflectionfollowedbyatranslation2. Reflectionfollowedbyarotation3. Rotationfollowedbyatranslation
20. Whencomposingisometrieswhichcompositionscanbereplacedbyasingleisometry?(JoeH.)
21. Sometimesacompositionoftworeflectionshasfixedpointsandsometimesitdoesn’t.Writeaparagraphinwhichyouidentifywhenithasfixedpointsandexplainwhythishappens.(RyanF.)
Chapter9(HyperbolicandSphericalGeometry)22. Answerifthefollowingquestionsaretruealways,sometimes,orneverin
hyperbolicgeometry.Justify.(Kim)1.ABisasuperparalleltolinel.ABistheclosestpossibleparalleltolinel.
2.Linejisperpendiculartolinek.Linekisperpendiculartolinel.Thereisalinemthatisperpendiculartobothlinesjandl.3.ConstructarbitrarytriangleABC.PointDislocatedonlineACandisbetweenpointsAandC.ThedefectoftriangleABDisgreaterthanthedefectoftriangleABC.
23. Prove:AAAisacongruencepostulateinHyperbolicGeometry.(Kaitlyn)24. ConstructSaccheriQuadrilateralABCDandthetriangleassociatedwithitwith
vertexEandintersectingthebaseatpointsFandG.FandGandconstructedtobemidpointsofAEandBErespectively.ProvetheanglesumofatriangleisequaltothesumofthesummitanglesofitsassociatedSaccheriQuadrilateral.(Lina)
25. TrueorFalse–providejustificationusingtheUpperhalfplanemodel.(Dan)a. Givenh‐lineAandpointB,notonA,thereexistsoneandonlyoneh‐line
paralleltoA.b. IfCandDaretwoparallelh‐lines,andDisparalleltoh‐lineEaswell,thenC
andEarealsoparallel.c. Givenanh‐lineAandapointB,notonA,thereexistsoneuniqueperpendicularthroughAtoB.
26. Statewhetherthefollowingstatementistrueorfalse:thelengthofonesideofanequilateraltriangledeterminesthemeasurementofoneangleinhyperbolicgeometry.Then,thelongerthelengthofasideofanequilateraltrianglethelargerthemeasurementofangleis.(Taki)
27. (KatyK.)
MAT211:FinalExamReviewStudent‐WrittenSolutionsFall2010
Chapter1(Intro)1. (KatySpencer)
2. (Maggie)
1.ConsidertheparallelogramABCDwithsideABparalleltosideDCandsideADparalleltosideBCanddiagonalsACandBD.2.Bythealternateinteriorangletheorem,weknowthatangleADMiscongruenttoangleCBMandalso,angleDAMiscongruenttoangleBCM.(AngleABMcongruenttoangleCDM.AngleDCMiscongruenttoangleBAM).3.Bythedefinitionofparallellines,weknowthatLinesareparalleliftheylieinthesameplane,andarethesamedistanceapartovertheirentirelength.Hence,weknowthatADiscongruenttoBCandABiscongruenttoDC.4.ByASA,weknowthattriangleAMDiscongruenttotriangleBMCandalsothattriangleAMBiscongruenttotriangleDMC.ByCPCTC,weknowthatlineAMiscongruenttolineMCandlineBMiscongruenttolineMD.5.Therefore,sincethesearecongruent,weknowthatthediagonalsofparallelogramABCDbisecteachother.
3. (Greg)
Finitegeometry(Fe‐Fo)4. (Ryan)Conjecture1:Noonelinecontainsallnpoints.
Proofbycontradiction:Assumeonelinedoescontainallnpoints.Sincen>2,thiscontradictsAxiom3,whichstates,“eachlinecontainsexactlytwopoints.”Thus,bydisprovingourassumption,weprovethat,“noonelinecontainsallnpoints.”Conjecture2:Thereexistexactly[1+2+…+(n‐1)]linesinthesystem.Conjecture3:Anyonepointliesonexactly(n‐1)lines.
5. (StephanieM.)Case1:Tryandprovethatforeverygroupofthreeathletes,therearelessthan10sports.Case2:Tryandprovethatforeverygroupofthreeathletes,therearemorethan10sports.Case3:Thereareexactlytensportstheyhaveshared.Case3:TheathletesareAdam(A),Billy(B),Cody(C),Drew(D),andEvan(E).ThissupportsAxiom1.Thepossiblegroupsofthreeare[ABC,ABD,ABE,ACD,ACE,ADE,BCD,BCE,BDE,andCDE].ThissupportsAxiom2.InorderforAxiom3tobecorrect,theremustbetendifferentsportsbecausenotalloftheathleteshaveplayedthesamesport.
6. (StephanieS.)Theorem1.Assumetwodistinctcollegesdonotcontainexactlyonestudent.Thisleadstotwocases.
Case1‐Twodistinctcollegescontainzerostudent.Butthisviolatesaxiom4,whichstates“Anytwodistinctcollegescontainatleastonestudentwhichbelongstoboth”
Case2‐Twodistinctcollegescontainmorethanonestudent.Weknowthattherearethreestudentsinthesystembyaxiom1.Soweonlyneedtoconsiderwhethertwodistinctcollegescontaintwostudentsorthreestudents.Butbyaxiom3,“Notallstudentsbelongtothesamecollege”weknowthattwodistinctcollegescannotcontainthreestudents.Andtwodistinctcollegescannotcontaintwostudents,becausethisviolatesaxiom2.Thus,becausebothcasesleadtocontradictionsoutoriginalassumptionmusthavebeenfalse.So,twodistinctcollegescontainexactlyonestudent.
Theorem2.Assumeacollegecancontainthreedistinctstudents.Thisleadstotwocases,becausebyaxiom1weknowthatthereareexactlythreestudentsandthreecolleges.
Case1‐allstudentsbelongtoonecollege,butthisviolatesaxiom3“Notallstudentsbelongtothesamecollege”Case2‐Eachcollegecontainsonestudent,thisviolatesaxiom2“Twodistinctstudentsbelongtoexactlyonecollege”Thus,sincebothcasesledtocontradictionsouroriginalassumptionmusthavebeenfalse.Thereacollegecannotcontainthreedistinctstudents.
Theorem3.Assumetheredoesnotexistasetoftwocollegesthatcontainallthestudentsofthesystem.Thisleadstotwocases.
Case1‐onecollegecontainsallstudents,thisviolatesaxiom3“Notallstudentsbelongtothesamecollege”Case2‐Everycollegecontainsjustonestudent.Thisviolatesaxiom2“Twodistinctstudentsbelongtoexactlyonecollege”andweknowthereareonlythreestudentsbecauseofaxiom1.
Thus,becausethetwocasesleadtocontradictionsandwithaxiom4“Anytwodistinctcollegescontainatleastonestudentwhichbelongstoboth”andsincethereareonlythreecollegesstatedbyaxiom1.Outoriginalassumptionisfalse.Thereforethereexistsasetoftwocollegesthatcontainallthestudentsofthesystem.Chapter2(Triangles)7. (Ellen)DrawtriangleABCwithincenterI.DropaperpendicularfromItoeach
sideofthetriangle.LabelthepointsatwhichtheseperpendicularsandsidesmeetasM,N,andP,asshowninthefigure.WeseektoprovethatMIiscongruenttoNIwhichiscongruenttoPI,becauseaperpendicularistheshortestdistancebetweenapointandaline(asshowninprevioushomework)andthereforeifMI,NI,andPIarecongruent,Iisequidistantfromeachside.
AngleAMIandangleAPIarerightanglesbecausetheyareformedbyperpendiculars.AngleAMIandangleAPIarecongruentbecauseallrightanglesarecongruent,accordingtoEuclid’s4thpostulate.AngleMAIandanglePAIarecongruentbecausetheyareformedbyananglebisector.AIisananglebisector
becausetheincenterisformed,bydefinition,bytheanglebisectorsofthetriangle.AIiscongruenttoAIbythereflexiveproperty.Therefore,byAAScongruent,triangleAMIandtriangleAPIarecongruent.ByCPCTC,MIiscongruenttoPI.Similarly,angleBMIandangleBNIarecongruentbecausetheyarerightangles,byEuclid’s4thpostulate.AngleMBIandangleNBIarecongruentbecausetheyareformedbyananglebisector.BIiscongruenttoBIbythereflexiveproperty.Therefore,triangleMBIandtriangleNBIarecongruentbyAAScongruence.ByCPCTC,MIiscongruenttoNI.Therefore,sinceMIiscongruenttoPIandMIiscongruenttoNI,bythetransitiveproperty,MI,NI,andPIarecongruenttoeachotherandsotheincenterisequidistantfromeverysideofthetriangle.
8. (Toni)1.True:Weshowedinclassthatthediagonalsofaparallelogramalwaysbisecteachother
Implicationi. Hypothesis:Aisaparallelogramii. Conclusion:Thediagonalsbisecteachotheriii. Converse:Ifthediagonalsofanobjectbisecteachotherthan
theobjectisaparallelogramiv. Contrapositive:Ifthediagonalsofanobjectdonotbisecteach
otherthantheobjectisnotaparallelogram2.False:Notallrighttrianglesareisosceles,meaningthattheycanhavetwoanglesthatdonotalwaysequaleachother
Implicationv. Hypothesis:Aisarighttrianglevi. Conclusion:Aisanisoscelestrianglevii. Converse:IfAisalwaysanisoscelestrianglethanAisaright
triangleviii. Contrapositive:IfAisnotanisoscelestrianglethanAisnota
righttriangle3.True:Weshowedinclassthatthisistrue
Notanimplication4.False:allsquaresarerhombibutnotallrhombiaresquares
Notanimplication5.False:everyrectangledoeshavethreesideBUTallrighttrianglesdonothavethreeanglesthatareequal(itisimpossibletohaveatrianglethathas3‐90degreeangles)
Notanimplication9. (Anna)WehavetriangleABCandtriangleDEF,suchthatangleAiscongruentto
angleD,sideABiscongruenttosideDEandangleBiscongruenttoangleE.AssumethatsideBCisnotcongruenttosideEF.ThentriangleABCisnot
congruenttotriangleDEF,ascongruenttriangleshavecorrespondingsidesandanglesthatarecongruent.ThereexistssomepointXthatisnotthesameaspointFonrayEFsuchthatEXiscongruenttoBC.XmusteitherbebetweenEandF,ornotbetweenEandFonrayEF.SinceBCiscongruenttoEX,angleBiscongruenttoangleE,andABiscongruenttoDE,triangleABCiscongruenttotriangleDEXbySAS.Bytrianglecongruence,angleBACiscongruenttoangleEDX.
Case1:XisbetweenEandFBytheangleadditionpostulate,angleEDX+angleXDFequalsangleEDF.However,sinceangleBAC=angleEDF,andangleEDX=angleBAC,wehavethatangleBAC+angleXDFequalsangleBAC,whichimpliesthatangleXDFhasameasureofofzerodegrees,butthisisnotthecase,sothereisacontradiction.Case2:XisnotbetweenEandF,butonrayEF.Bytheangleadditionpostulate,angleEDF+angleFDXequalsangleEDX.SinceangleBAC=angleEDFandangleBAC=angleEDX,angleBAC+angleFDXequalsangleBAC,soangleFDXhasameasureofzerodegrees.SinceFisnotthesamepointasX,FDXhasanonzeromeasure,sothisisacontradiction.Becausebothpotentialcasesresultedincontradictions,itmustbethatBCiscongruenttoEF.Therefore,bySAS,triangleABCiscongruenttotriangleDEF.ThismeansthatiftwotriangleshavetheconditionsofASA,thentheymustbecongruent.
10. (Ketty)
Chapter3(Circles)11. (Clarissa)
12. (Mike)
13. (Caitlin)Proof:WeneedtoshowthatrayXRbisectsAngleYXZ.Drop
perpendicularsfromRtolinesXY,YZ,XZ,andlabelthefeetoftheseperpendicularsD,E,andFrespectively.TriangleRDXandTriangleRFXarerighttrianglesbyconstruction.RX=RXbythereflexiveproperty.TriangleRDY=TriangleREYbyAAS.TriangleRFZ=TriangleREZ,alsobyAAS.RD=RE,andRE=Rf,sincetheyarecorrespondingsidesofcongruenttriangles.Thus,RD=RF
bythetransitiveproperty.Therefore,TriangleRDX=TriangleRFXbyHL.SowecanconcludethatAngleYXR=AngleZXR,whichmeansthatXRbisectsAngleYXZ.
14. (JoeK.)
Chapter4(Analyticgeometry)15. (Adam) Proof: Plot the points A(0,0), B(0,a), and C(b,0). Let the slope of segment
AB be called m1. Using the slope formula, m1= (a-0)/(0-0) = a/0, which is undefined. So segment AB is vertical. Let the slope of segment AC be called m2. Using the slope formula, m2 = (0-0)/(b-0) = 0/b = 0. So segment AC is horizontal. Since AB is vertical and AC is horizontal, AB is perpendicular to AC and angle A is a right angle. Therefore, triangle ABC is a right triangle. Using the distance formula,
, , and
. So by substitution ,
and . Therefore, , which is the Pythagorean theorem.
16. (Kevin)
17. (KatieSkerik)
18. (Jamie)a.)ConstructarighttriangleABC,sothatpointBisattheorigin(0,0).HavepointAbeonthey‐axisandpointCbeonthex‐axis,sothattheyhavethecoordinatesA(0,2a)andC(2c,0).Seethediagramaboveforavisual.Weknowthat∠ABCisarightanglebecauseitisformedbythelinesBCandBAand:TheslopeofBCis(0‐0/0‐2c)=0TheslopeofBAis(0‐2a/0‐0)=UndefinedSincelineswithaslopeofzeroareperpendiculartolineswithanundefinedslope,weknowBCisperpendiculartoBA.Therefore,theyformarightangleattheirintersection,so∠ABCisarightangle.Therefore,thistrianglecanbeanyrighttriangle,aslongasAisonthey‐axisandCisonthex‐axisandBistheonlypointattheorigin.b.)Usingthemidpointformula,wecanfindthemidpointofthehypotenuseAC:((0+2c/2),(2a+0/2))=(c,a).WecancallthismidpointpointM.
Chapter6(Transformationalgeometry)19. (Justin)
1.Orientation:reverse;Fixedpoint:None2.Orientation:reverse;Fixedpoint:Ifthepointofrotationisonthelineofreflectionitwillbeafixedpoint.3.Orientation:same;Fixedpoint:Thereisapointintheplanesuchthatthiscompositioncouldhavebeenjustasinglerotationaboutapoint.
20. (JoeH.)Atranslation‐translationcompositioncanbereplacedbyatranslation.Areflection‐reflectioncompositionoverinterestinglinescanbereplacedbyareflection.Areflection‐reflectioncompositionoverparallellinescanbereplacedbyatranslation.Atranslation‐rotationcompositioncanbereplacedbyarotation.
21. (RyanF.)There are two cases in which a composition of two reflections will have fixed points. Case1) If the lines of reflection intersect, the point at which they intersect would be fixed because we know that with a single reflection the points on the line of reflection are fixed, then the common point between the two lines will also be fixed. Case2) When the second reflection is reflected over the same line as the first reflection, the only fixed points would be those contained on the line of reflection. It is the same as saying that the two lines infinitely intersect and like in case1, where they intersect will not be reflected, they will be fixed.
Chapter9(HyperbolicandSphericalGeometry)22. (Kim)1.Falsebydefinition.Asuperparallelisanyraythatisnotalimiting
parallelray,whichgoesthroughapointPandisparalleltol.BecausealimitingparallelisaraythatoriginatesatpointPandistheclosestpossibleraytoalinelwithoutintersectingit,alimitingparallelraywouldbetheanswertothisquestion,notsuperparallel.
2.Sometimes.Thisstatementistruewhenlinemisconcurrentwithlinek.Thisstatementisfalsewhenlinemisnotconcurrentwithlinek.Thisisbecauseinhyperbolicgeometryrectanglesdonotexist.Rectanglesdonotexistbecausealltrianglesinhyperbolichaveapositivedefect.IfwedrawlinesegmentACwecanseethisastwotriangles.Sincetrianglesmusthaveapositivedefecttheiranglesumisalwayslessthan180,andneverequalto180.Since
defectsareadditivewecanseethisalsoasaddingthesumsofthetwotriangles.Thesumofthisquadrilateralwouldbea+b+c+d+e+f.Weknowthatanglebis90˚sinceitisarightangle.Thisisalsotruefordand(c+e).Thuswehave90+90+90+a+f.Wealsoknowthatthesetwotrianglestogethermustbelessthan360(sumoftwo
trianglesthatarelessthan180).Thus360>270+�+�and90>�+�.ThereforetheangleDABcannotbearightangleandthusthereisnotasecondcommonparallelbetweenlinejandl.
3.ThedefectoftriangleABDis180‐a‐b‐f.ThedefectoftriangleBDCis180‐c‐d‐e.ThedefectoftriangleABCisthesumofthedefectsofABDandBCD.ThuswehavethedefectoftriangleABCis180‐a‐b‐f+180‐c‐d‐e.Fromherewecanseethattheanglesofonetriangleplustheanglesofasecondtrianglewillbelargerthaneitheroftheindividualtriangles.ThustheanglesoftriangleABDsumtoasmallernumberthantheanglesoftriangleABC.Thusif
a+b+f<a+b+c+dthenthedefectsofthesetriangleswillbe180‐a‐b‐ffortriangleABDand180‐a‐b‐c‐dfortriangleABC.Ifwesettheseequationsequaltoeachotherfornowwithaquestionovertheequalssign(inthiscasea‘doesnotequal’sign)weget180‐a‐b‐c‐d≠180‐a‐b‐f.Doingsomemanipulationwegeta+b+c+d≠a+b+f.Bylookingatourearlierequationwecanreplacethisdoesnotequalsigntoagreaterthansignanda+b+c+d>a+b+f.Therefore,bygoingbackwardswecanseethat180‐a‐b‐c‐d<180‐a‐b‐f,andthusthedefectoftriangleABDisgreaterthanthedefectoftriangleABC.23. (Kaitlyn)
Inthediagramabove,triangleADEandtriangleABCareconstructedsuchthatangleEDAiscongruenttoangleCBAandangleBCAiscongruenttoangleDEA.AngleAiscongruenttoangleAbythereflexiveproperty.InHyperbolicGeometry,theangledefectisfoundbysubtractingthesumoftheanglesfrom180degrees.Usingthisformula,thedefectoftriangleABC=180–m<A–m<CBA–m<BCA.Similarly,thedefectoftriangleAED=180–m<A–m<EDA–m<DEA.Theangledefectofaquadrilateralis360degreesminusthesumoftheangles,whichmeansthatthedefectofquadrilateralBCED=360–m<CBA–m<BCA–m<CED–m<BDE.Angledefectsareadditive,whichmeansthatthedefectoftriangleABCisequaltothesumofthedefectsoftriangleADEandquadrilateralBCED.Thus,defectoftriangleABC=180–m<A–m<EDA–m<DEA+360–m<CBA–m<BCA–m<CED–m<BDE.Aftercombiningliketerms,thedefectoftriangleABC=540–m<A–(m<DEA+m<CED)–(m<EDA+m<BDE)–m<CBA–m<BCA.Since<AECand<ADBareastraightangles,m<AEC=180degrees=m<ADB.<AECand<ADBaresplitintotwoanglesbysegmentED,whichmeansthatm<AEC=m<DEA+m<CEDandm<ADB=m<EDA+m<BDE.Bythetransitiveproperty,m<DEA+m<CED=180degrees=m<EDA+m<BDE.Bysubstitution,thedefectoftriangleABC=540–m<A–(180)–(180)–m<CBA–m<BCA=180–m<A–m<CBA–m<BCA,whichshowsthatthedefectoftriangleADE+thedefectofquadrilateralBCED=thedefectoftriangleABC.ThismeansthatthedefectoftriangleABCislargerthan
thedefectoftriangleADE,whichinturnmeansthatthesumoftheanglesoftriangleABCmustbesmallerthanthesumoftheanglesoftriangleADE.Thisshowsthatalargertrianglecannothavethesameanglemeasureasasmallertriangle.Similarly,asmallertrianglehasalargeranglemeasurethanalargertriangle.Thus,theonlywaythattwotrianglescanhavethesameanglemeasureiswhenthesidesarealsocongruent.
24. (Lina)ConstructaperpendicularfromEtosegmentCDwithintersectionpoint,R.WeknowthemeasureofangleGREis90degreesbyconstruction.WealsoknowthatmeasureofangleFREis90degreesbyconstruction.SincethebaseanglesofaSaccheriQuadrilateralarerightangles,weknowangleBDGiscongruenttoangleERGsinceallrightanglesarecongruent.WeknowBGiscongruenttoEGbyconstruction.WeknowangleRGEiscongruenttoangleDGBsinceverticalanglesarecongruent;thus,byHAtriangleGREiscongruenttotriangleGDB.ByCPCTC,angleREGiscongruenttoangleGBD.Similarly,weknowthatangleACFiscongruenttoangleFREsinceallrightanglesarecongruent.WeknowAFiscongruenttoEFbyconstruction.AngleAFCiscongruenttoangleEFRsinceverticalanglesarecongruent.ByHAtriangleACFiscongruenttotriangleERF.ByCPCTC,angleCAFiscongruenttoangleFEF.Thesumofthesummitanglesisequaltothem<CAF+m<EAB+m<ABE+m<EBD,bysubstitution,thesumisequaltom<FER+m<EAB+m<ABE+m<BERwhichisthesumofmeasuresoftheanglesoftheassociatedtriangle.
25. (Dan)
26. (Taki)Theanswerisfalse.Thefigureisacounterexampleofthestatement.
27. (KatyK.)
