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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.
Let BF cut ED and AC at H and GObserve that H and G are harmonic conjugate points of F and BSo GF/GB=HF/HBBut S3/S= GF/GB and S2/S3= HF/HBThe result will follow.
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)*SS3 = (AF/AD) * ACD = (AF/AD) * (CD/BC) * SS / S3 = (AD / AF) * (BC / CD)S2 / ADE = DF / ADADE / S1 = AE / BE => ADE = (AE / BE) * S1S2 = (DF/AD) * ADE = (FD / AD) * (AE / BE) * S1S1 / 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 / S3Therefore S1 / S2 = S / S3Hence S.S2 = S1.S3