In a parallelogram ABCD, any point E is taken on the side BC.

Join BM and AC. 

$\operatorname{ar}(\Delta \mathrm{ADC})=\frac{1}{2} b h=\frac{1}{2} \times \mathrm{DC} \times h$

$\operatorname{ar}(\Delta \mathrm{ABM})=\frac{1}{2} \times \mathrm{AB} \times h$

AB = DC                               (Since ABCD is a parallelogram)
So, ar(∆ADC) = ar(∆ABM)
⇒">⇒ar(∆ADC) + ar(∆AMC) = ar(∆ABM) + ar(∆AMC)
⇒">⇒ar(∆ADM) = ar(ABMC)
Hence Proved