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 $\displaystyle \mathrm{EL}=2 \mathrm{BL}$.

\[\begin{array}{l} \Delta B L C \sim \Delta E L A \\ \frac{B L}{E L}=\frac{B C}{E A} \ldots \ldots \ldots \end{array} \]In $\displaystyle \triangle E A B$ M is mid point of BE $\displaystyle \mathrm{DM} \| A B$ \[\begin{array}{l} \therefore A D=D E \\ \Rightarrow A E=2 A D \\ \Rightarrow A E=2 B C . \end{array} \] From (i) and (ii) EL=2BL