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.