$20 \cdot 48$ In circle $O$, chord $AB = AC$, chord $AD$ intersects $BC$ at $E$, if $AC = 12$, $AE = 8$, then $AD$ equals
(A) 27.
(B) 24.
(C) 21.
(D) 20.
(E) 18.
(9th American High School Mathematics Examination, 1958)
[Solution] Connect $B D$. By the property of inscribed angles, it is easy to prove
$$
\begin{array}{l} 
\triangle A B E \backsim \triangle A D B, \\
\therefore \quad A B^{2}=A E \cdot A D,
\end{array}
$$

then
$$
A D=\frac{12^{2}}{8}=18 \text {. }
$$

Therefore, the answer is $(E)$.