$ABCD$ is a parallelogram. $A$ circle through $A$ and $B$ is drawn such that it intersects $AD$ at $P$ and $BC$ at $Q$. Prove that $P, Q, C,$ and $D$ are concyclic.

(N/A) $ABCD$ is a parallelogram. $A$ circle passes through $A$ and $B$,intersecting $AD$ at $P$ and $BC$ at $Q$. We need to prove that $P, Q, C,$ and $D$ are concyclic.
$1$. Since $ABQP$ is a cyclic quadrilateral,the exterior angle formed by extending side $AP$ to $D$ is equal to the interior opposite angle.
Let $\angle QPD$ be the exterior angle at $P$. Then,$\angle QPD = \angle ABQ$.
$2$. In parallelogram $ABCD$,$AD \parallel BC$. Since $AB$ is a transversal,$\angle A + \angle B = 180^{\circ}$. However,considering the parallel lines $AB$ and $DC$ with transversal $BC$,we have $\angle B + \angle C = 180^{\circ}$ (consecutive interior angles).
$3$. Since $\angle QPD = \angle ABQ$ (exterior angle property of cyclic quadrilateral $ABQP$),and we know $\angle ABQ + \angle C = 180^{\circ}$,we can substitute $\angle QPD$ for $\angle ABQ$.
$4$. Therefore,$\angle QPD + \angle C = 180^{\circ}$.
$5$. Since the sum of the opposite angles of quadrilateral $PDCQ$ is $180^{\circ}$,$PDCQ$ is a cyclic quadrilateral.
Hence,the points $P, Q, C,$ and $D$ are concyclic.