In trapezium $PQRS, PQ || RS$ and diagonals $PR$ and $QS$ intersect at point $M$. Prove that,$ar(PQS) = ar(QPR)$.

(N/A) Given: In trapezium $PQRS$,$PQ || RS$.
To prove: $ar(PQS) = ar(QPR)$.
Proof:
$1$. Triangles $PQS$ and $QPR$ are on the same base $PQ$.
$2$. Since $PQ || RS$,the distance between the parallel lines $PQ$ and $RS$ is constant.
$3$. Therefore,the height of triangle $PQS$ (with base $PQ$) and the height of triangle $QPR$ (with base $PQ$) are equal to the distance between the parallel lines $PQ$ and $RS$.
$4$. We know that the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
$5$. Since both triangles have the same base $PQ$ and the same height (the distance between $PQ$ and $RS$),their areas must be equal.
$6$. Thus,$ar(PQS) = ar(QPR)$.