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).
Solution:
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