Question 6
O is any point on the diagonal PR of a parallelogram PQRS (figure). Prove that ar ( $Δ$ PSO) = ar ( $Δ$ PQO).

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Solution

Construction: Join SQ which intersects PR at B .

Proof:
We know that diagonals of a parallelogram bisect each other. So, B is the midpoint of SQ.
Here, PB is a median of $Δ$ QPS and we know that, a median of a triangle divides it into two triangles of equal area.
$∴$ ar( $Δ$ BPQ) = ar ( $Δ$ BPS) ...(i)
ar( $Δ$ OBQ) = ar ( $Δ$ OBS) ....(ii)
On adding Eqs. (i) and (ii), we get,
$a r (Δ B P Q) + a r (Δ O B Q) = a r (Δ B P S) + a r (Δ O B S)$
$\Rightarrow a r (Δ P Q O) = a r (Δ P S O)$

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