Proposed Problem
Click the figure below to see the complete problem 312.
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Complete Problem 312
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Monday, June 29, 2009
Problem 312: Cyclic Quadrilateral, Side extensions, Tangents
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Given GE=a,HF=b,EF=c and ABCD is cyclic so angles A+c=B+D=180 and angle EBC=angleEDA=D,angleECB=angleEAD=A=angleFCD, angleFDC=angleFBA=B and angleBCD=angleECF=c=180-A by sine rule for tr EBCCB/CE=Sin(A+D)/Sin D and CD/CF = Sin(A+B)/Sin B, and EH,FH are tangents so a^2 =EB.EA=EB^2+EB.AB=EC^2+EC.CD and b^2 =FD^2+FD.AD=FC^2+FC.CB now by cosine rule for tr CEF, c^2=CE^2+CF^2-2.CE.CF.Cos C therefore c^2= a^2+b^2-2CE.CF Cos C-EC.CD-FC.CB= a^2+b^2-CE.CF(2Cos C- CD/CF - CB/CE) by appling values for CD/CF and CB/CE we will get the value of (2Cos C - CD/CF -CB/CE) =0 so c^2=a^2+b^2-o hence c^2=a^2+b^2
ReplyDeletegiven GE=a,HF=b,EF=c and ABCD is cyclic so A+C=B+D=180 and angleEBC=angleEDA=D,angleECB=angleEAD=A=angleFCD,angleFDC=angleFBA=B, angleBCD=angleECF=180-A, by sine rule tr EBC, CB/EC= Sin(A+D)/Sin D simillarly CD/CF=Sin(A+B)/Sin B, and EG,FH are tangents so a^2=EB.EA=EB^2+EB.AB=EC^2+EC.CD andb^2=FD^2+FD.AD=FC^2+FC.CB and by cosine rule for trCEF, c^2=CE^2+CF^2-2CE.CF.Cos C=a^2-EC.CD+b^2-FC.CB-2.CE.CF.Cos C=a^2+b^2-CE.CF(2CosC-CD/CF-CB/CE)=a^2+b^2-CE.CF(0)=a^2+b^2
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ReplyDeleteGiven GE=a,HF=b,EF=c, and ABCD is cyclic so A+C=B+D=180 and angleEBC=angleEDA=D,angleECB=angleEAD=A=angleFCD,angleFDC=angleFBA=Band angleBCD=angleECF=C=180-A, we have by sine rulefor triangles CB/EC=Sin(A+D)/SinD, CD/CF=Sin(A+B)/SinB=-Sin(A-D)/SinD, so CB/EC + CD/CF=(2CosA.SinD)/SinD=2CosA so CB.CF+CD.CE=2EC.CFCos(180-C) and b^2-FC^2+a^2-CE^2=-2EC.CFCosC hence a^2+b^2=FC^2+EC^2-2EC.CF.CosC so a^2+b^2=c^2
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