Question

In a parallelogram ABCD, any point E is taken on the side BC. AE and DC when produced meet at a point M. Prove that
ar(∆ADM) = ar(ABMC)

Answer 1

Join BM and AC.
ar(∆ADC) = $\frac{1}{2}bh$ = $\frac{1}{2}\times \mathrm{DC}\times h$
ar(∆ABM) = $\frac{1}{2}\times \mathrm{AB}\times h$
AB = DC (Since ABCD is a parallelogram)
So, ar(∆ADC) = ar(∆ABM)
$\Rightarrow$ar(∆ADC) + ar(∆AMC) = ar(∆ABM) + ar(∆AMC)
$\Rightarrow$ar(∆ADM) = ar(ABMC)
Hence Proved