In parallelogram $PQRS$,the bisector of $\angle P$ intersects $RS$ at $M$ and the bisector of $\angle R$ intersects $PQ$ at $N$. Prove that $PNRM$ is a parallelogram.

(N/A) $1$. In parallelogram $PQRS$,we have $\angle P = \angle R$ (opposite angles are equal) and $PQ \parallel RS$.
$2$. Let the bisector of $\angle P$ be $PM$ and the bisector of $\angle R$ be $RN$.
$3$. Since $PM$ bisects $\angle P$,$\angle SPM = \angle RPM = \frac{1}{2} \angle P$.
$4$. Since $RN$ bisects $\angle R$,$\angle QRN = \angle PRN = \frac{1}{2} \angle R$.
$5$. Since $\angle P = \angle R$,it follows that $\frac{1}{2} \angle P = \frac{1}{2} \angle R$,so $\angle RPM = \angle PRN$.
$6$. These are alternate interior angles for lines $PM$ and $RN$ with transversal $PR$. Therefore,$PM \parallel RN$.
$7$. Also,$PQ \parallel RS$ implies $PN \parallel RM$ (since $N$ lies on $PQ$ and $M$ lies on $RS$).
$8$. Since both pairs of opposite sides are parallel ($PM \parallel RN$ and $PN \parallel RM$),$PNRM$ is a parallelogram.