$20 \cdot 58 P$ is a point on the plane where $\triangle A B C$ is located, and $P A = P B$, also $\angle A P B = 2 \angle A C B$, and $A C$ intersects $B P$ at point $D$. If $P B = 3, P D = 2$. Then $A D \cdot A C$ equals
(A) 5 .
(B) 6 .
(C) 7 .
(D) 8 .
(E) 9 .
[Solution] Construct the circumcircle of $\triangle A B C$, then $P$ is precisely the center of this circle, and $P A$ is its radius.
Extend $B P$ to intersect $\odot P$ at point $E$. By the intersecting chords theorem, we have
$$
\begin{aligned}
A D \cdot D C & =B D \cdot E D=B D \cdot(P D+P E) \\
& =(3-2)(3+2)=5 .
\end{aligned}
$$

Therefore, the answer is $(A)$.