6. As shown in the figure, in the right trapezoid $A B C D$, the base $A B=13$, $C D=8, A D \perp A B$, and $A D=$ 12. Then the distance from $A$ to the side $B C$ is ( ).
(A) 12
(B) 13
(C) 10
(D) $\frac{12 \times 21}{13}$
6. (A).

Draw $C C^{\vee} \perp A B$, then
$$
B C=13-8=5 \text {. }
$$

Let the distance from $A$ to $B C$ be $h$,
Connect AC.
$$
\begin{aligned}
& \because S_{\triangle A C D}+S_{\triangle A B C} \\
& =S_{\text {UEABCD }}, \\
\therefore & \frac{1}{2} \times 8 \times 12+\frac{1}{2} \times h \times 13=\frac{1}{2}(8+13) \times 12
\end{aligned}
$$

Therefore, $h=12$.