Thursday, July 9, 2015

Geometry Problem 1131: Triangle Area, Two Cevians, Equal Product of Areas

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1131: Triangle Area, Two Cevians, Equal Product of Areas.

2 comments:

  1. Let BF cut ED and AC at H and G
    Observe that H and G are harmonic conjugate points of F and B
    So GF/GB=HF/HB
    But S3/S= GF/GB and S2/S3= HF/HB
    The result will follow.

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  2. Lets compute the ratios S/S3 and S1/S2 and prove they are equal:

    S3/ACD = AF/AD (triangles with same height)
    ACD/S = CD/BC => ACD = (CD/BC)*S
    S3 = (AF/AD) * ACD = (AF/AD) * (CD/BC) * S
    S / S3 = (AD / AF) * (BC / CD)

    S2 / ADE = DF / AD
    ADE / S1 = AE / BE => ADE = (AE / BE) * S1
    S2 = (DF/AD) * ADE = (FD / AD) * (AE / BE) * S1
    S1 / S2 = (AD / DF) * (BE / AE)

    Next use Menelaus' theorem for the triangle ABD and the line CFE:

    ( BE / AE) * ( AF / DF ) * ( CD / BC ) = 1
    ( BE / AE) * ( AF / DF ) = BC / CD (multiply both sides by AD/AF)
    ( BE / AE) * ( AD / DF ) = (BC / CD) * (AD / AF)

    These are exactly the expressions above for S1 / S2 and S / S3
    Therefore S1 / S2 = S / S3

    Hence S.S2 = S1.S3


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