Proposed Problem

Click the figure below to see the complete problem 391 about Triangle, Parallel lines, Collinear points.

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Complete Problem 391

Level: High School, SAT Prep, College geometry

## Sunday, November 15, 2009

### Problem 391. Triangle, Parallel lines, Collinear points

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Let m=distance AD , n=distance DC and c= distance AB

ReplyDeleteWe have FA/FB= m/n ( Tri. AFD ~ Tri. ABC)

So FB= cn/(m+n) ( FA+FB=c)

EC/EB= n/m ( Tri CED~ Tri. CBA)

EB/GF=AB/AF= (m+n)/n

We have KB/KA= (m+n)/m ( see comment of the Problem 390)

So KB= c*(m+n)/(2m+n)

1 Calculate KF=KB-FB= c*m^2/((2m+n)*(m+n))

So KF/KB=m^2/(m+n)^2

2 Calculate MC/MF =EC/GF ( Tri. FMG~ Tri. CME)

=(n/m)* EB/GF

Replace EB/GF=(m+n)/n we get MC/MF= n*(m+n)/m^2

3 JB/JC= (m+n)/n ( see comment of the Problem 390)

4 Consider Triangle FBC and 3 points K, M and J

Verify that (JB/JC)*(MC/MF)*(KF/KB)=

=(m+n)/n *n*(m+n)/m^2 * m^2/(m+n)^2 =1

So 3 points K, M and L are collinear per Menelaus’s theorem

Peter Tran

vstran@yahoo.com