Parallelograms on the same base and between the same parallels are equal in area.

**Given : **Two parallelograms ABCD and ABEF are on the same base AB and betweem the same parallels.

**To prove : **

**Proof : **In parallelograms ABEF and ABCD, one common figure can be observed, which is trapezium ABCF.

Both the parallelograms consist of two figures : a triangle and a trapezium. Now, the trapezium is the common figure between these two parallelograms.

We can write :

And

The right hand sides of thew above two equations will be equal only when ar (BCE) = ar (AFD.

For this, we have to prove that BCE and AFD are congruent triangles.

Let us prove it.

In parallelogram ABCD,

AB = AD

Similarly, in parallelogram ABEF,

AB = EF

From (3) and (4), we get

Or

Now, in

............... (Corresponding angles)

................ (Proved above)

................. (Corresponding angles)

.............. (ASA)

.......... (6)

Using (1), (2) and (6), we get ar (ABEF) = ar (ABCD).