Question

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

Answer 1

We know
ar(∆APB) = $\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$ .....(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,
ar(∆BQC) = $\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)$ .....(2)
From (1) and (2)
ar(∆APB) = ar(∆BQC)
Hence Proved