$ABCD$ 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: $APCQ$ is a parallelogram.
(N/A) In quadrilateral $APCQ,$
$AP \parallel QC$ (since $AB \parallel CD$ and $P, Q$ are mid-points of $AB$ and $CD$ respectively) ............ $(1)$
$AP = \frac{1}{2} AB$ and $CQ = \frac{1}{2} CD$ (Given that $P$ and $Q$ are mid-points)
Also,$AB = CD$ (Opposite sides of a parallelogram are equal)
Since $AB = CD$,then $\frac{1}{2} AB = \frac{1}{2} CD$,which implies $AP = QC$ ............ $(2)$
From $(1)$ and $(2)$,since one pair of opposite sides is equal and parallel,$APCQ$ is a parallelogram.
$AP \parallel QC$ (since $AB \parallel CD$ and $P, Q$ are mid-points of $AB$ and $CD$ respectively) ............ $(1)$
$AP = \frac{1}{2} AB$ and $CQ = \frac{1}{2} CD$ (Given that $P$ and $Q$ are mid-points)
Also,$AB = CD$ (Opposite sides of a parallelogram are equal)
Since $AB = CD$,then $\frac{1}{2} AB = \frac{1}{2} CD$,which implies $AP = QC$ ............ $(2)$
From $(1)$ and $(2)$,since one pair of opposite sides is equal and parallel,$APCQ$ is a parallelogram.
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