Question

In a parallelogram ABCD, P is any point on the side AB. A line AQ drawn parallel to PC intersects DC at Q. BD intersects AQ at S and PC at R. Prove that ar(PRB)=ar(DSQ).

Answer 1

In ΔPRB and ΔDSQ
As taking a general case let Q and P be the mid-points of the parallelogram, as nothing is given.
So DQ = PB
And AQ|| PC (given)
∠PRB =∠CRS (vertically opposite angle)
∠CRS = ∠QSD (corresponding angles)
So ∠PRB = ∠QSD

And ∠PBR = ∠QDS

So ΔPRB is congruent to  ΔDSQ (By AAS)
So Ar(PRB) = Ar(DSQ) (Hence proved)