Question

Question 3
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).

Answer 1

It can be observed that $\Delta BQC$ and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.
$\therefore Area\left(\Delta BQC\right)=\frac{1}{2}Area\left(ABCD\right)...\left(1\right)$
Similarly, $\Delta APB$ and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.
$\therefore Area\left(\Delta APB\right)=\frac{1}{2}Area\left(ABCD\right)...\left(2\right)$
From equation (1) and (2), we obtain;
$Area\left(\Delta BQC\right)=Area\left(\Delta APB\right)$