Proposed Problem

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

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Complete Problem 392

Level: High School, SAT Prep, College geometry

## Sunday, November 15, 2009

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

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Let m= distance AD and n= distance DC

ReplyDeleteWe 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