In the given figure, ABCD is a parallelogram. BE bisects CD at M and intersects AC at L . Prove that EL = 2BL .

$\triangle ALE \sim \triangle CLB$
$\Rightarrow \frac{AL}{CL} = \frac{EL}{BL}$
quad (i)
Also $\triangle CLM \sim \triangle ALB$
$\Rightarrow \frac{AL}{CL} = \frac{AB}{CM}$
$\Rightarrow \frac{AL}{CL} = \frac{CD}{CM}$
quad $\{AB = CD\}$
quad (ii)
Using (i) and (ii)
$\frac{EL}{BL} = \frac{2CM}{CM}$
$\Rightarrow EL = 2BL$