$20 \cdot 57$ As shown in the figure, it is known that $A D$ is the diameter of $\odot O$, and $A D^{\prime} \perp$ $B C$, and $A B, A C$ intersect the circle at $E, F$ respectively. Then the following equation that must hold is
(A) $A E \cdot B E=A F \cdot C F$.
(B) $A E \cdot A B=A O \cdot A D^{\prime}$.
(C) $A E \cdot A B=A F \cdot A C$.
(D) $A E \cdot A F=A O \cdot A D$.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1998)
[Solution] Connect $E D$, it is easy to prove that Rt $\triangle A E D \backsim R t \triangle A D^{\prime} B$, then we have
$$
\frac{A E}{A D^{\prime}}=\frac{A D^{\prime}}{A B} \text {, i.e., } A E \cdot A B=A D \cdot A D^{\prime} \text {. }
$$

Connect $F D$, similarly we can prove $A F \cdot A C=A D \cdot A D^{\prime}$.
$$
\therefore \quad A E \cdot A B=A F \cdot A C \text {. }
$$

Therefore, the answer is (C).