See complete Problem 131
Van Aubel Theorem, Concurrent Cevians, Sum of Ratios. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Wednesday, June 25, 2008
Elearn Geometry Problem 131
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ReplyDeleteBP/PE=[ABCP]/[APC]=[ABP]/[APC]+[PBC]/[APC]=BD/DC+BF/FA
ReplyDeleteSolution to problem 131.
ReplyDeleteLet’s use Menelaus’ theorem to triangle ABE, with the transversal FPC, and triangle CBE, with transversal DPA, getting
(AF/FB).(BP/PE).(EC/CA) = 1
and (CD/DB).(BP/PE).(EC/CA) = 1.
Thus BF/FA = (BP/PE).(EC/CA)
nd BC/DC = (BP/PE).(EA/CA).
Adding the last equalities,
BF/FA + BC/DC = (BP/PE).(EC/CA) + (BP/PE).(EA/CA) = BP(EC+EA)/(PE.CA) = (BP.CA)/(PE.CA) = BP/PE.