9. As shown in Figure 3, in $\triangle A B C$, $A D$ intersects side $B C$ at point $D, \angle B=45^{\circ}, \angle A D C=60^{\circ}, D C=2 B D$. Then $\angle C$ equals $\qquad$
$9.75^{\circ}$.
As shown in Figure 7, draw $C E \perp A D$ at point $E$, and connect $B E$.

Since $\angle A D C=60^{\circ}$, therefore, $\angle E C D=30^{\circ}$, and
$$
D C=2 D E .
$$

Also, $D C=2 B D$, so
$$
\begin{aligned}
D E & =B D, \\
\angle D B E & =\angle D E B=30^{\circ} .
\end{aligned}
$$

Thus, $\angle D B E=\angle E C B=30^{\circ}, C E=B E$.
Since $\angle A B D=45^{\circ}$, therefore,
$$
\angle E B A=\angle E A B=15^{\circ}, A E=B E .
$$

Hence, $A E=C E$, $\triangle A C E$ is an isosceles right triangle, and $\angle A C E=45^{\circ}$.
Therefore, $\angle A C D=\angle D C E+\angle A C E=75^{\circ}$.