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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.)