Prove the Pythagorean theorem using analytic geometry. Be ...
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MAT 211: Final Exam Review Student‐Written Questions Fall 2010 Chapter 1 (Intro) 1. A triangle is isosceles if and only if the base angles are congruent. (Katy Spencer) 2. Prove that the diagonals of a parallelogram bisect one another (relies on Euclid’s 5 th Postulate) (Maggie) 3. A quadrilateral is said to be cyclic if there is a circle passing through all its four vertices. Given a quadrilateral is cyclic, prove that the sum of the measures of its non‐adjacent interior angles is 180 degrees. (Greg) Finite geometry (Fe‐Fo) 4. Given the following set of axioms, write three conjectures and prove one of them. (Ryan D.) Axiom 1: There exist exactly n points in the system. Axiom 2: Every set of two points lies on exactly one line. Axiom 3: Each line contains exactly two points. 5. Prove the conjecture from the given axioms. (Stephanie M.) A1: There is a group of five athletes. A2: For every three athletes, there is a common sport they have shared. A3: Not all of the athletes have played the same sport. Conjecture: There are 10 possible sports that have been shared. 6. There are many different theorems that can be proved from the Axioms below. Prove one of the theorems provided. (Stephanie S.) Axiom 1. There exist exactly three distinct students, and three distinct colleges. Axiom 2. Two distinct students belong to exactly one college. Axiom 3. Not all students belong to the same college. Axiom 4. Any two distinct colleges contain at least one student which belongs to both. Theorem 1. Two distinct colleges contain exactly one student. Theorem 2. A college cannot contain three distinct students. Theorem 3. There exists a set of two colleges that contain all the students. Chapter 2 (Triangles) 7. Consider the triangle ABC with incenter I. Prove that the incenter is a point that is equidistant from each side of the triangle. (Ellen) 8. Determine whether the following statements are true or false. If the statement is an implication state the hypothesis and conclusion. Then state the converse and the contrapositive. (Toni) 1. If A is a parallelogram than the diagonals bisect each other. 2. If A is a right triangle than it is always an isosceles triangle 3. The center of a circle lies on the perpendicular bisector of two points 4. All rhombi are squares 5. Every rectangle has three sides, and all right triangles are equiangular. 9. Assuming that SAS is true, prove ASA. (Anna)
Transcript of Prove the Pythagorean theorem using analytic geometry. Be ...
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.)
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