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