$P$ and $Q$ are the mid-points of the opposite sides $AB$ and $CD$ of a parallelogram $ABCD.$ $AQ$ intersects $DP$ at $S$ and $BQ$ intersects $CP$ at $R.$ Show that $PRQS$ is a parallelogram.

(N/A) Given: $A$ parallelogram $ABCD$ in which $P$ and $Q$ are the mid-points of the sides $AB$ and $CD$ respectively. $AQ$ intersects $DP$ at $S$ and $BQ$ intersects $CP$ at $R$.
To prove: $PRQS$ is a parallelogram.
Proof: Since $ABCD$ is a parallelogram,$DC \parallel AB$ and $DC = AB$.
Since $P$ is the mid-point of $AB$ and $Q$ is the mid-point of $CD$,we have $AP = \frac{1}{2} AB$ and $QC = \frac{1}{2} CD$.
Since $AB = CD$,it follows that $AP = QC$.
Also,$AP \parallel QC$ because $AB \parallel DC$.
Since one pair of opposite sides ($AP$ and $QC$) is equal and parallel,$APCQ$ is a parallelogram.
Therefore,$AQ \parallel PC$,which implies $SQ \parallel PR$.
Similarly,we can show that $APQD$ is a parallelogram (since $AP \parallel DQ$ and $AP = DQ$),which implies $AQ \parallel DP$. Also,$PBCQ$ is a parallelogram (since $PB \parallel QC$ and $PB = QC$),which implies $BQ \parallel CP$.
In quadrilateral $PRQS$,we have $SQ \parallel PR$ (from $AQ \parallel PC$) and $SP \parallel QR$ (from $DP \parallel BQ$).
Since both pairs of opposite sides are parallel,$PRQS$ is a parallelogram.