$O$ is any point on the diagonal $PR$ of a parallelogram $PQRS$. Prove that $\operatorname{ar}(\triangle PSO) = \operatorname{ar}(\triangle PQO)$.

(N/A) Let the diagonals $PR$ and $SQ$ of the parallelogram $PQRS$ intersect at point $M$.
Since the diagonals of a parallelogram bisect each other,$M$ is the midpoint of $PR$ and $SQ$. Thus,$SM = MQ$.
In $\triangle PQS$,$PM$ is the median because $M$ is the midpoint of $SQ$.
Since a median of a triangle divides it into two triangles of equal area,we have:
$\operatorname{ar}(\triangle PSM) = \operatorname{ar}(\triangle PQM) \quad \dots(1)$
Similarly,in $\triangle OQS$,$OM$ is the median because $M$ is the midpoint of $SQ$.
Therefore,$\operatorname{ar}(\triangle OSM) = \operatorname{ar}(\triangle OQM) \quad \dots(2)$
Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\triangle PSM) + \operatorname{ar}(\triangle OSM) = \operatorname{ar}(\triangle PQM) + \operatorname{ar}(\triangle OQM)$
$\operatorname{ar}(\triangle PSO) = \operatorname{ar}(\triangle PQO)$
Hence,it is proved.