In the figure,$X$ and $Y$ are respectively the mid-points of the opposite sides $AD$ and $BC$ of a parallelogram $ABCD$. Also,$BX$ and $DY$ intersect $AC$ at $P$ and $Q$ respectively. Show that $AP = PQ = QC$.

(N/A) Given: $ABCD$ is a parallelogram,$X$ is the mid-point of $AD$,and $Y$ is the mid-point of $BC$.
$1$. Since $AD \parallel BC$ and $AD = BC$,we have $DX \parallel BY$ and $DX = BY$ (as $X$ and $Y$ are mid-points).
$2$. Therefore,$XBYD$ is a parallelogram because one pair of opposite sides is equal and parallel.
$3$. This implies $PX \parallel QD$ and $PY \parallel BQ$.
$4$. In $\triangle AQD$,$X$ is the mid-point of $AD$ and $XP \parallel QD$. By the Converse of Mid-point Theorem,$P$ is the mid-point of $AQ$. Thus,$AP = PQ$.
$5$. In $\triangle CPB$,$Y$ is the mid-point of $BC$ and $YQ \parallel PB$. By the Converse of Mid-point Theorem,$Q$ is the mid-point of $PC$. Thus,$PQ = QC$.
$6$. From the above two results,we get $AP = PQ = QC$.