In figure, ABCD and AEFD are two parallelograms. Prove that `ar (DeltaPEA) = ar (DeltaQFD)`.

Explanation

Join AC.
Now, `triangleADC and triangleADQ` being on the same base AD and between the same parallels AD and BC, are equal in area.

`therefore ar(triangleADC)=ar(triangleADQ)`
`rArr ar(triangleADC)-ar(triangleADP)=ar(triangleADQ)-ar(triangleADP)`
`rArr ar(triangleAPC)=ar(triangleDPQ)." "...(i)`
Also, `triangle APC and triangleBPC` being on the same base PC and between the same parallels PC and AB, are equal in area.
`therefore ar(triangleAPC)=ar(triangleBPC)." "...(ii)`
From (i) and (ii), we get `ar(triangleBPC)=ar(triangleDPQ)`.