Sunday, November 15, 2009

Problem 391. Triangle, Parallel lines, Collinear points

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

 Problem 391. Triangle, Parallel lines, Collinear points.
See also:
Complete Problem 391


Level: High School, SAT Prep, College geometry

1 comment:

  1. Let m=distance AD , n=distance DC and c= distance AB
    We 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

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