Two circles intersect at two points $B$ and $C$. Through $B$,two line segments $ABD$ and $PBQ$ are drawn to intersect the circles at $A, D$ and $P, Q$ respectively (see figure). Prove that $\angle ACP = \angle QCD$.

(N/A) Since angles in the same segment of a circle are equal:
$1$. In the circle on the left,$\angle ACP = \angle ABP$ (angles subtended by the same arc $AP$).
$2$. In the circle on the right,$\angle QCD = \angle QBD$ (angles subtended by the same arc $QD$).
$3$. Since $ABD$ and $PBQ$ are straight lines intersecting at $B$,$\angle ABP$ and $\angle QBD$ are vertically opposite angles.
$4$. Therefore,$\angle ABP = \angle QBD$.
$5$. From steps $1$,$2$,and $4$,we conclude that $\angle ACP = \angle QCD$.