Geometry Problem

Click the figure below to see the complete problem 479 about Problem 479: Triangle, Cevians, Concurrency, Transversal, Proportion.

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Complete Problem 479

Level: High School, SAT Prep, College geometry

## Saturday, July 24, 2010

### Problem 479: Triangle, Cevians, Concurrency, Transversal, Proportion

Labels:
cevian,
concurrent,
proportions,
transversal,
triangle

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join B to G

ReplyDeletein ▲ABG from ceva's theorem

BD/DA∙AG/CG = BE/EC => AG/CG = BE/EC∙DA/BD (1)

in ▲ABC from ceva's theorem

BD/DA∙AF/FC∙EC/BE = 1 (2)

multiply (1) to (2)

AG/CG = (BE/EC∙DA/BD)∙(BD/DA∙AF/FC∙EC/BE)

AG/CG = AF/FC

Below is the proving of 2nd part of Problem 479

ReplyDelete1. Apply Ceva’s theorem to triangle DBE and point of concurrence O . We have:

(HD/HE)*(CE/CB)*(AB/AD)=1 so HD/HE= (CB/CE)*(AD/AB)

2. Apply Menelaus’s theorem for triangle DBE and secant ACG . We have:

(GE/GD)*(AD/AB)*(CB/CE)=1 so GD/GE=(AD/AB)*(CB/CE)

3. From the result of steps 1 and 2 we get HD/HE=GD/GE

Peter Tran

vstran@yahoo.com