Question

In a parallelogram ABCD,any point E is taken on the sides BC.AE and DC when produced meet at a point M.Prove that $ar\left(△ADM\right)=ar\left(ABMC\right)$.

Answer 1

Join BM and AC.
ar(∆ADC) = 1/2bh = 1/2×DC×h
ar(∆ABM) = 1/2×AB×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