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Similarity Question - Asked by a student in Yahoo forum

ABCD is a parallelogram. M is the mid point of CD. The line BM is drawn intersecting AC at L and AD produced at E. Prove that EL=2BL

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In triangle LMC & ALB 

these are similar because   MLC =  BLA

&  LMC =  LBA

so they are similar by angle-angle  theorem.

now we will have

MC/AB= LC/AL= ML/LB = 1/2

because MC/CD=1/2  & CD=AB

so MC/AB=1/2

now we will prove similarity for the triangles  AEL & triangle  BLC



so by angle angle  theorem two triangles  are similar

so we will have the ratio of sides as


we already have AL/LC = 2/1

so EL / LB = 2 /1

hence proved.

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