Sunday, November 15, 2009

Problem 392. Triangle, Parallel lines, Collinear points, Concurrent lines

Proposed Problem
Click the figure below to see the complete problem 392 about Triangle, Parallel lines, Collinear points, Concurrent lines.

See also:
Complete Problem 392

Level: High School, SAT Prep, College geometry

1 comment:

1. Let m= distance AD and n= distance DC
We have HE/HD=FB/FA = n/ m [ Tri. AFD ~ Tri. ABC]
GD/GF=DE/FA=FB/FA =n/m [ Tri. AFG~ Tri. EDG]
JE/JB= n^2/(m+n)^2 [ See comment from problem 391]
KB/KF= (m+n)^2/m^2 [ See comment from problem 391]
Let GH cut EF at point P and JK cut EF at point P’.
1. Consider triangle EFD and secant HGP. Apply Menelaus theorem we have
(HE/HD) *(GD/GF) * ( PF/PE) =1
Replace values of HE/HD and GD/GF from above to this equation
We get PF/PE= m^2/n^2
2 Consider Triangle BEF and secant JKP’ . Apply Menelaus theorem we have
(JE/JB) *(KB/KF) *( P’F/P’E)= 1
Replace values of JE/JB and KB/KF from above to this equation
We get P’F/P’E= m^2/n^2
3 From step 1 and 2 we can conclude that P coincides with P’ ( both P and P’ located outside segment EF)
And EF, JK and HG are concurrent

Peter Tran
vstran@yahoo.com