Points $P$ and $Q$ have been taken on opposite sides $AB$ and $CD$ respectively of a parallelogram $ABCD$ such that $AP = CQ$ (see figure). Show that $AC$ and $PQ$ bisect each other.

(N/A) Given: $ABCD$ is a parallelogram,$P$ is on $AB$,$Q$ is on $CD$,and $AP = CQ$.
Let $AC$ and $PQ$ intersect at point $O$.
In $\Delta OAP$ and $\Delta OCQ$:
$1$. $AP = CQ$ (Given)
$2$. $\angle OAP = \angle OCQ$ (Alternate interior angles,as $AB \parallel CD$)
$3$. $\angle AOP = \angle COQ$ (Vertically opposite angles)
Therefore,$\Delta OAP \cong \Delta OCQ$ by $ASA$ congruence rule.
By $CPCT$,$OA = OC$ and $OP = OQ$.
Since $OA = OC$,$O$ is the midpoint of $AC$.
Since $OP = OQ$,$O$ is the midpoint of $PQ$.
Thus,$AC$ and $PQ$ bisect each other.