Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD…

In $\triangle BMC$ and $\triangle EMD$
MC = MD
$\angle CMB = \angle EMD$
$\angle MBC = \angle MED$
$\therefore \triangle BMC \cong \triangle EMD$
$\Rightarrow$ BC = DE
But AD = BC
$\therefore$ AD = DE
$\Rightarrow$ AE = $2$ BC
In $\triangle AEL \sim \triangle CBL$
$\therefore \frac{EL}{BL} = \frac{AE}{BC}$
$\frac{EL}{BL} = \frac{2BC}{BC}$
$\frac{EL}{BL} = 2$
$\Rightarrow EL = 2 BL$