**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 ∠ ALE= ∠ BLC ∠ EAL = ∠ BCL so by angle angle theorem two triangles are similar so we will have the ratio of sides as AE/BC= AL/LC = EL/LB we already have AL/LC = 2/1 so EL / LB = 2 /1 hence proved. |

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