Question

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

Answer 1

Construct lines AC and BM.

Let us take h as the distance between AB and CD

We know that

Area of △ ACD = ½ × CD × h

In the same way

Area of △ ABM = ½ × AB × h

From the figure we know that AB = CD

So it can be written as

Area of △ ABM = ½ × CD × h

So we get

Area of △ ABM = Area of △ ACD

Let us add △ ACM on both sides

Area of △ ABM + Area of △ ACM = Area of △ ACD + Area of △ ACM

So we get

Area of ABMC = Area of △ ADM

Therefore, it is proved that ar (△ ADM) = ar (ABMC).