M is the mid point of …

In {tex}\Delta{/tex}BMC and {tex}\Delta{/tex} EMD, we have
MC = MD [{tex}\because{/tex} M is the mid-point of CD]
{tex}\angle \mathrm { CMB } = \angle E M D{/tex} [Vertically opposite angles]
and, {tex}\angle M B C = \angle M E D{/tex} [Alternate angles]
So, by AAS-criterion of congruence, we have
{tex}\therefore \quad \Delta B M C \cong \Delta E M D{/tex}
{tex}\Rightarrow{/tex} BC = DE .......(i)
Also, AD = BC [{tex}\because{/tex} ABCD is a parallelogram] ...... (ii)

Adding (i) and (ii),we get,
AD + DE = BC + BC
{tex}\Rightarrow{/tex} AE = 2 BC ...(iii)
Now, in {tex}\Delta{/tex}AEL and {tex}\Delta{/tex}CBL, we have
{tex}\angle A L E = \angle C L B{/tex} [Vertically opposite angles]
{tex}\angle E A L = \angle B C L{/tex} [Alternate angles]
So, by AA-criterion of similarity of triangles, we have

{tex}\Delta A E L \sim \Delta C B L{/tex}
{tex}\Rightarrow \quad \frac { E L } { B L } = \frac { A E } { C B }{/tex}
{tex}\Rightarrow \quad \frac { E L } { B L } = \frac { 2 B C } { B C }{/tex} [Using equations (iii)]
{tex}\Rightarrow \quad \frac { E L } { B L } = 2{/tex}
{tex}\Rightarrow{/tex} EL = 2 BL