Saturday, July 17, 2010

Problem 474: Parallelogram, Diagonal, Circle, Vertex

Geometry Problem
Click the figure below to see the complete problem 474 about Parallelogram, Diagonal, Circle, Vertex.

Problem 474: Parallelogram, Diagonal, Circle, Vertex
See also:
Complete Problem 474

Level: High School, SAT Prep, College geometry

3 comments:

  1. Let BG intersect AC at I . We have IB=IC and IA=IC
    Let circumcircles of Triangles DFG and AEG intersect AC at N and M
    Connect GE and GF
    1. Note that Angle(GFC)=Angle(GND) both angle supplement to angle(NGF)
    Angle(CEG)=Angle(BMC)
    2. Angle(BMI)=Angle(IND) both angle supplement to angle(GND)
    And BM//DN

    3. We have Tri. BMI congruence with Tri. NDI ( case ASA) so IM=IN
    4. We have CE.CB+CF.CD=CG.CN+CG.CM= CG*(CN+CM)
    5. Replace CN+CN=CA we get the result


    Peter Tran
    vstran@yahoo.com

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  2. I'm INDSHAMAT from University of Kelaniya,Srilanka.This is my solution....
    Angle ACD=Angle CAB(ABCD is a parallelogram)
    Angle ACD=Angle GEF and Angle ACB=Angle EFG(GFCE is a cyclic quadrilateral)
    Hence GEF and ABC triangles are similar.
    So EF/CA=GE/AB=GF/CB,Therefore EF=CA.GF/BC(1) and GE=AB.GF/CB(2)
    From Ptolemy's theorem CG.EF=CE.GF+GE.FC
    Replacing from (1)&(2)for EF and GE we have
    (CG.CA.GF)/CB=CE.GF+(AB.GF.FC)/CB
    Excluding GF from both sides, CA.CG=CB.CE+CD.CF

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  3. There are some typo errors in my previous comment. Below is my correction.
    Let BD intersect AC at I . We have IB=ID and IA=IC
    Let circumcircles of triangles DFG and BEG intersect AC at N and M respectively
    Connect GE and GF
    1. Note that Angle(GFC)=Angle(GND) both angle supplement to angle(GFD)
    Angle(CEG)=Angle(BMC)
    2. Angle(BMI)=Angle(IND) both angle supplement to angle(GND)
    And BM//DN

    3. We have Tri. BMI congruence with Tri. NDI ( case ASA) so IM=IN
    4. We have CE.CB+CF.CD=CG.CN+CG.CM= CG*(CN+CM)
    5. Replace CN+CN=CA we get the result

    ReplyDelete