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

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Solution

In $△ B M C$ and $△ E M D$
$∠ B M C = ∠ E M D$ [Vertically opposite]
$M C = D M$ [Given]
$∠ B C M = ∠ E D M$ [Alternate angles]
$∴△ B M C ≅△ E M D$ [By ASA]
Hence, $B C = D E$ [By CPCT] $\to (1)$
$A E = A D + D E = B C + B C = 2 B C$ $\to (2)$
Now, $△ B L C \sim△ E L A$ (AA Similarly)
$\frac{B L}{E L} = \frac{B C}{A E}$ [By CPCT]
$\frac{B L}{E L} = \frac{B C}{2 B C}$
$\frac{B L}{E L} = \frac{1}{2}$
$E L = 2 B L$

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