$ABCD$ is a parallelogram in which $P$ and $Q$ are mid-points of opposite sides $AB$ and $CD$ (see Fig.). If $AQ$ intersects $DP$ at $S$ and $BQ$ intersects $CP$ at $R$,show that $PSQR$ is a parallelogram.
(N/A) Given: $ABCD$ is a parallelogram,$P$ is the mid-point of $AB$,and $Q$ is the mid-point of $CD$.
Since $AB \,||\, CD$ and $AB = CD$,we have $AP \,||\, QC$ and $AP = \frac{1}{2} AB = \frac{1}{2} CD = QC$.
Since $AP \,||\, QC$ and $AP = QC$,$APCQ$ is a parallelogram.
Therefore,$AQ \,||\, PC$,which implies $SQ \,||\, PR$.
Similarly,$PB \,||\, DQ$ and $PB = \frac{1}{2} AB = \frac{1}{2} CD = DQ$.
Since $PB \,||\, DQ$ and $PB = DQ$,$PBQD$ is a parallelogram.
Therefore,$DP \,||\, BQ$,which implies $SP \,||\, QR$.
Since both pairs of opposite sides are parallel ($SQ \,||\, PR$ and $SP \,||\, QR$),$PSQR$ is a parallelogram.
Since $AB \,||\, CD$ and $AB = CD$,we have $AP \,||\, QC$ and $AP = \frac{1}{2} AB = \frac{1}{2} CD = QC$.
Since $AP \,||\, QC$ and $AP = QC$,$APCQ$ is a parallelogram.
Therefore,$AQ \,||\, PC$,which implies $SQ \,||\, PR$.
Similarly,$PB \,||\, DQ$ and $PB = \frac{1}{2} AB = \frac{1}{2} CD = DQ$.
Since $PB \,||\, DQ$ and $PB = DQ$,$PBQD$ is a parallelogram.
Therefore,$DP \,||\, BQ$,which implies $SP \,||\, QR$.
Since both pairs of opposite sides are parallel ($SQ \,||\, PR$ and $SP \,||\, QR$),$PSQR$ is a parallelogram.
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