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
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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