Proposed Problem
Click the figure below to see the complete problem 391 about Triangle, Parallel lines, Collinear points.
See also:
Complete Problem 391
Level: High School, SAT Prep, College geometry
Sunday, November 15, 2009
Problem 391. Triangle, Parallel lines, Collinear points
Subscribe to:
Post Comments (Atom)
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