9. $120^{\circ}-\alpha$.

From the given, we may assume
$$
\begin{array}{l}
\angle D A C=\angle D C A=\beta, \\
\angle C D B=\angle C B D=\gamma . \\
\text { Then } \angle A D C=180^{\circ}-2 \beta, \angle D C B=180^{\circ}-2 \gamma \text {. } \\
\text { By } \angle A D B+\angle A C B \\
=\left(180^{\circ}-2 \beta-\gamma\right)+\left(180^{\circ}-2 \gamma-\beta\right)=180^{\circ} \\
\Rightarrow \beta+\gamma=60^{\circ} \text {. } \\
\text { Therefore } \angle D A B+\angle C B A \\
=360^{\circ}-\angle A D C-\angle D C B \\
=2(\beta+\gamma)=120^{\circ} \\
\Rightarrow \angle C B A=120^{\circ}-\alpha .
\end{array}
$$
9. As shown in Figure 2, in quadrilateral $A B C D$, $A D=D C=$ $C B$, and $\angle A D B+\angle A C B=180^{\circ}$, given $\angle D A B$ $=\alpha$. Then $\angle C B A=$ $\qquad$