Proposed Problem

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

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Complete Problem 390

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Sunday, November 15, 2009

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

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Extend AH to BC at J

ReplyDeleteExtend 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)

AF/AB=AD/AC= m/(m+n) ( Tri. ADF ~ Tri. ACB)

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