$O$ is any point on the diagonal $P R$ of parallelogram $P Q R S$ .
Prove that $a r (△ P S O) = a r (△ P Q O)$ .

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Solution

To prove:
$a r (△ P S O) = a r (△ P Q O)$
Construction: Join $S Q$ , bisect the diagonal $P R$ at $M$ .

Proof:
Since diagonals of a parallelogram bisect each other, so $S M = M Q$

Therefore, $P M$ is a median of $Δ P Q S$

$a r (Δ P S M) = a r (Δ P Q M)$ ......(1)

[ $∵$ Median divides a triangle into two triangles of equal area]

Refer image,

Again, as $O M$ is the median of $Δ O S Q$ , so

$a r (Δ O S M) = a r (Δ O Q M)$ ..........(2)

Adding (1) and (2), we get

$a r (Δ P S M) + a r (Δ O S M) =$ $a r (Δ P Q M) + a r (Δ O Q M)$

$\Rightarrow a r (Δ P S O) = a r (P Q O)$

Hence, proved.

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