Question

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

Answer 1

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 $\Delta$ QPS and we know that, a median of a triangle divides it into two triangles of equal area.
$\therefore$ ar($\Delta$ BPQ) = ar ($\Delta$BPS) ...(i)
ar($\Delta$ OBQ) = ar ($\Delta$OBS) ....(ii)
On adding Eqs. (i) and (ii), we get,
$ar\left(\Delta BPQ\right)+ar\left(\Delta OBQ\right)=ar\left(\Delta BPS\right)+ar\left(\Delta OBS\right)$
$\Rightarrow ar\left(\Delta PQO\right)=ar\left(\Delta PSO\right)$