## Sunday, November 15, 2009

### Problem 390. Triangle, Parallel lines, Collinear points

Proposed Problem
Click the figure below to see the complete problem 390 about Triangle, Parallel lines, Collinear points.

Complete Problem 390
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

#### 1 comment:

1. Extend AH to BC at J
Extend CG to AB at K
Let m=distance AD and n=distance DC
We have HC/HF=DC/DA=n/m ( Tri. CHD ~ Tri. CFA)
CB/CE=CA/CD= (m+n)/n ( Tri. CED~Tri. CBA)
GE/GA=DC/DA=n/m ( Tri. ADG ~ Tri ACE)

1. Apply Menelaus theorem on Tri. FBC with secant AHJ
(JB/JC)*(HC/HF)*(AF/AB)=1
Replace HC/HF= n/m and AF/AB= m/(m+n)
We get JB/JC= (m+n)/n
2. Apply Menelaus theorem on Tri. ABE with secant KGC
(KA/KB)*(CB/CE)*(GE/GA)=1
Replace CB/CE=(m+n)/n and GE/GA= n/m
We get KA/KB= m/(m+n_
3 Consider Tri. ABC and cevians BD, AJ and CK
Calculate (JB/JC) * (DC/DA)*(KA/KB)=
(m+n)/n * n/m *m/(m+n) = 1
So these 3 cevians will concurrence at M per Ceva’s theorem and B, M and D are collinear.

Peter Tran
vstran@yahoo.com