Monday, June 29, 2009

Problem 312: Cyclic Quadrilateral, Side extensions, Tangents

Proposed Problem
Click the figure below to see the complete problem 312.

Problem 312: Cyclic Quadrilateral, Side extensions, Tangents.
See also:
Complete Problem 312
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 4:25 AM
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7 comments:

  1. vijay9290009015November 18, 2009 at 10:49 PM

    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

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  2. vijay9290009015December 4, 2009 at 1:51 AM

    given 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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  3. vijay9290009015December 4, 2009 at 11:38 PM

    Given 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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  4. Antonio GutierrezAugust 26, 2013 at 9:37 PM

    Problem 312, Suggestion for a synthetic solution: Problem 915 at www.gogeometry.com/school-college/p915-triangle-circle-cyclic-quadrilateral-concyclic.htm

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  5. AjitAugust 26, 2013 at 10:13 PM

    By problem # 915 there exists a point I on EF such that quads FDCI and EBCI are both comcyclic. Now EG^2 = EB*EA = EC*ED = EI*EF. ///ly, FH^2 = FD*FA = FC*FB = FI*EF. Hence, EG^2 + FH^2 = EI*EF + FI*EF = (EI + FI)EF = EF^2 or b^2 + c^2 = a^2.

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  6. Peter TranAugust 27, 2013 at 8:48 AM

    http://img9.imageshack.us/img9/6039/ugqa.png
    Let circumcircle of triangle CDF cut EF at M
    Since ABCD and CDFM are cyclic quadrilaterals
    It easy to show that angle(BCM) supplement to angle (BEM) => quadrilateral BCME is cyclic.
    Since E is located on radical line or circles O and CDFM => a^2=EC.ED=EM.EF
    Since F is located on radical line or circles O and BCME=> b^2=FC.FB=FM.FE
    Add above expressions side by side we have a^2+b^2= c^2

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