If P and Q are any two points lying respectively on the sides DC and AD

We know

$\operatorname{ar}(\Delta \mathrm{APB})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$       .....(1)      

  [If a triangle and a parallelogram are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram]

Similarly, 

$\operatorname{ar}(\Delta \mathrm{BQC})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$       .....(2)

From (1) and (2)
ar(∆APB) = ar(∆BQC)
Hence Proved