P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

Triangle ABP and parallelogram ABCD lie on the same base AB and between the same parallels, AB ∥ CD
Hence, ar (APB) = (1/2) ar (ABCD) ….(1)
[∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.]
Similarly,
Triangle BQC and Parallelogram ABCD lie on the same base AB and between same parallels, AD ∥ BC.
Hence, ar(BQC = (1/2)ar(ABCD) …(2)
From the equation (1) and (2), ar (APB) = ar(BQC).