18.104 As shown in the figure, in the right trapezoid, the base $AB=$
13, $CD=8, AD \perp AB$, and $AD=12$, then the distance from $A$ to $BC$ is
(A) 12.
(B) 13.
(C) 10.
(D) $\frac{12 \cdot 21}{13}$.
(E) $\frac{13+8}{2}$.
(China Junior High School Mathematics Correspondence Competition, 1985)
[Solution] Draw $C F \perp A B$ at $F$, we can find that $B C=A B$.
Thus,
$\triangle A B E \cong \triangle C B F$.
Therefore,
$$
A E=C F=12 \text {. }
$$

So the answer is $(A)$.