Answer :

**Construction:** Draw FG perpendicular to AB.

**Proof:** We have,

*BE* = 2 *EA*

And,

*DF =* 2*FC*

*AB - AE =*2 *AE*

And,

*DC - FC = 2 FC*

*AB = 3 AE*

And,

*DC =* 3 *FC*

*AE =* *AB* and *FC =* *DC* (i)

But,

AB = DC

Then,

*AE* = *FC* (Opposite sides of a parallelogram)

Thus,

*AE* || *FC* such that *AE = FC*

Then,

*AECF* is a parallelogram

Now, Area of parallelogram (AECF) = (AB × FG) [From (i)

3 Area of parallelogram AECF = AB × FG (ii)

And,

Area of parallelogram ABCD = AB × FG (iii)

Compare equation (ii) and (iii), we get

3 Area of parallelogram AECF = Area of parallelogram ABCD

Area of parallelogram AECF = Area of parallelogram ABCD

**Hence, proved**

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