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

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Step 1: Parallelogram ABCD with M as midpoint of CD, line BM extended intersecting AC at L and AD produced at E
Step 2: In $	riangle BMC$ and $	riangle EMD$: $MC = MD$, $\angle CMB = \angle EMD$ (vertically opposite), $\angle MBC = \angle MED$ (alternate angles). $	herefore 	riangle BMC \cong 	riangle EMD$. $\Rightarrow BC = DE$
Step 3: But $AD = BC$, $	herefore AD = DE$. $\Rightarrow AE = 2BC$
Step 4: $	riangle AEL \sim 	riangle CBL$ (AA similarity). $	herefore \frac{EL}{BL} = \frac{AE}{BC}$
Step 5: $\Rightarrow \frac{EL}{BL} = \frac{2BC}{BC} = 2$
Step 6: $\Rightarrow EL = 2BL$

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