KöMaL 2022 C1740

$P$ is an interior point of side $CD$ of a parallelogram $ABCD$, and $Q$ is an interior point of side $AB$ parallel to $CD$. The intersection of line segments $PA$ and $QD$ is $M$, and the intersection of line segments $PB$ and $QC$ is $N$. Assume that $MN\nparallel{AB}$, and $MN$ intersects the line of $CD$ at point $X$, and the line of $AB$ at point $Y$. Prove that $DX=BY$.