3. As shown in the figure, in $\triangle A B C$, $D$ is a point on $A B$, and $D C=$ $\frac{A D \cdot B C}{A C}$. Then ().
(A) $\triangle A D C \sim \triangle A C B$
(B) $\triangle B D C \sim \triangle B C A$
(C) $\triangle A D C \sim \triangle C D B$
(D) There are no similar triangles in the figure
3. (A).

From the problem, we know $\frac{D C}{A D}=\frac{B C}{A C}$,
$$
\text { and } \frac{D C}{A D}=\frac{\sin A}{\sin \angle A C D}, \frac{B C}{A C}=\frac{\sin A}{\sin B} \text {, }
$$

then $\sin \angle A C D=\sin B$.
But $\angle B+\angle A C D$
$$
<\angle B+\angle A C B<180^{\circ} \text {, }
$$

then $\angle A C D=\angle B$,
$\therefore \triangle A D C \backsim \triangle A C B$.