$P$ and $Q$ are points on opposite sides $AD$ and $BC$ of a parallelogram $ABCD$ such that $PQ$ passes through the point of intersection $O$ of its diagonals $AC$ and $BD$. Show that $PQ$ is bisected at $O$.

(N/A) $ABCD$ is a parallelogram. Its diagonals $AC$ and $BD$ bisect each other at $O$. $PQ$ passes through the point of intersection $O$ of its diagonals $AC$ and $BD$.
In $\triangle AOP$ and $\triangle COQ$,we have:
$\angle 3 = \angle 4$ [Alternate interior angles,as $AD \parallel BC$]
$OA = OC$ [Diagonals of a parallelogram bisect each other]
$\angle 1 = \angle 2$ [Vertically opposite angles]
Therefore,$\triangle AOP \cong \triangle COQ$ [By $ASA$ congruence rule]
So,$OP = OQ$ [$CPCT$]
Hence,$PQ$ is bisected at $O$.