In the given figure, join PG.
G is the mid-point of CD.
Thus, PG is a median of Δ DPC and it divides the triangle into parts of equal areas.
Also, we know that; if a parallelogram and a triangle lie on the same base and between the same parallels, then the area of the triangle is equal to half of the area of the parallelogram.
Here, parallelogram EFGD and DPG lie on the same base DG and between the same parallels DG and EF.
So, ar(ΔDPG=12ar(EFGD) ....(ii)
From equations (i) and (ii), we get,