In quadrilateral $ABCD$,$P$ is the midpoint of $\overline{BC}$. If $\overline{DP}$ produced intersects $\overline{AB}$ produced at $Q$,prove that $AB = 2CD$ given that $ABCD$ is a parallelogram.

(A) Given: $ABCD$ is a parallelogram,$P$ is the midpoint of $\overline{BC}$,and $\overline{DP}$ produced meets $\overline{AB}$ produced at $Q$.
To prove: $AB = 2CD$.
Proof:
$1$. Consider $\triangle DCP$ and $\triangle QBP$.
$2$. $\angle DCP = \angle QBP$ (Alternate interior angles,as $AB \parallel CD$).
$3$. $CP = BP$ (Given,$P$ is the midpoint of $BC$).
$4$. $\angle DPC = \angle QPB$ (Vertically opposite angles).
$5$. Therefore,$\triangle DCP \cong \triangle QBP$ by the $ASA$ congruence criterion.
$6$. By $CPCT$,$CD = BQ$.
$7$. Since $ABCD$ is a parallelogram,$AB = CD$ and $AB \parallel CD$.
$8$. From the figure,$AQ = AB + BQ$.
$9$. Since $BQ = CD$ and $CD = AB$,we have $AQ = AB + AB = 2AB$ (This specific problem usually implies $ABCD$ is a parallelogram where $AB=CD$).
$10$. If the condition $AB=2CD$ is to be proven,it typically relates to specific geometric configurations where $Q$ is defined such that $AB=BQ$.