We have,

Given:

There are more than 1 parallelogram, and their bases can be taken as common and they are between same parallels.

To Prove:

These parallelograms whose bases are same and are between the same parallel sides have equal area.

Proof:

Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.

Here,

AB ∥ DC and AE ∥ BF

We can represent area of parallelogram ABCD as,

…(i)

Now, area of parallelogram ABFE can be represented as,

Area of parallelogram ABFE

[∵ in right-angled ∆ADE, ]

⇒Area of parallelogram ABFE

[∵ , where k is scalar; is parallel to and hence ]

[∵ a scalar term can be taken out of a vector product]

[∵ ]

⇒Area of parallelogram ABFE …(ii)

From equation (i) and (ii), we can conclude that

Area of parallelogram ABCD = Area of parallelogram ABFE

Thus, parallelogram on same base and between same parallels are equal in area.

Hence, proved.