$20 \cdot 28$ In the circle $O$, $BC$ is the diameter of the circle, and $AB \perp BC$. Also, $ADOE$ is a straight line, $AP=AD$, and the length of $AB$ is twice the radius $R$. Then
(A) $AP^2=PB \cdot AB$.
(B) $AP \cdot DO=PB \cdot AD$.
(C) $AB^2=AD \cdot DE$.
(D) $AB \cdot AD=OB \cdot AO$.
(E) None of the above.
[Solution] Given that $AB$ is a tangent to $\odot O$, by the secant-tangent theorem, we have
$$
\begin{aligned}
AB^{2} & =AD \cdot AE=AD \cdot(AD+DE)=AD(AD+2R) \\
& =AD(AD+AB)=AD^{2}+AD \cdot AB,
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
AD^{2} & =AB^{2}-AD \cdot AB=AB(AB-AD) \\
& =AB(AB-AP)=AB \cdot PB .
\end{aligned}
$$

Thus, the correct choice is $(A)$.