Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$, angle $APB$ is twice angle $ACB$, and $\overline{AC}$ intersects $\overline{BP}$ at point $D$. If $PB = 3$ and $PD= 2$, then $AD\cdot CD =$

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$
The product of two lengths with a common point brings to mind the [Power of a Point Theorem](https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem).
Since $PA = PB$, we can make a circle with radius $PA$ that is centered on $P$, and both $A$ and $B$ will be on that circle.  Since $\angle APB = \widehat {AB} = 2 \angle ACB$, we can see that point $C$ will also lie on the circle, since the measure of arc $\widehat {AB}$ is twice the measure of inscribed angle $\angle ACB$, which is true for all inscribed angles.

Since $PDB$ is a line, we have $PD + DB = PB$, which gives $3 = DB + 2$, or $DB = 1$.
We now extend radius $PB = 3$ to diameter $EB = 6$.  Since $EDB$ is a line, we have $ED + DB = EB$, which gives $ED + 1 = 6$, or $ED = 5$.
Finally, we apply the power of a point theorem to point $D$.  This states that $AD \cdot DC = DB \cdot DE$.  Since $DB = 1$ and $DE = 5$, the desired product is $5$, which is $\boxed{A}$.