$ABCD$ is a parallelogram in which $P$ and $Q$ are mid-points of opposite sides $AB$ and $CD$ respectively. If $AQ$ intersects $DP$ at $S$ and $BQ$ intersects $CP$ at $R$,show that $DPBQ$ is a parallelogram.

(N/A) Given that $ABCD$ is a parallelogram,so $AB \,|| \,DC$ and $AB = DC$.
Since $P$ and $Q$ are mid-points of $AB$ and $CD$ respectively,we have $PB = \frac{1}{2} AB$ and $DQ = \frac{1}{2} DC$.
Since $AB = DC$,it follows that $\frac{1}{2} AB = \frac{1}{2} DC$,which implies $PB = DQ$.
Also,since $AB \,|| \,DC$,we have $PB \,|| \,DQ$.
In quadrilateral $DPBQ$,we have one pair of opposite sides ($PB$ and $DQ$) equal and parallel.
Therefore,$DPBQ$ is a parallelogram.