9. In the triangular prism $A B C-A_{1} B_{1} C_{1}$, $A C=\sqrt{2} A A_{1}=A_{1} C=1 . A B=\frac{\sqrt{6}}{3}, A_{1} B=$ $\frac{\sqrt{3}}{3}, B C=\frac{2 \sqrt{3}}{3}$, then $\angle B_{1} C_{1} C=$ $\qquad$ .
9. $\arccos \frac{\sqrt{3}}{6}$
$$
\begin{array}{c}
A_{1} B^{2}+A B^{2}=A A_{1}^{2} \Rightarrow \angle A_{1} B A=\frac{\pi}{2} \\
A B^{2}+B C^{2}=A C^{2} \Rightarrow \angle A B C=\frac{\pi}{2} \\
A_{1} B^{2}+A_{1} C^{2}=B C^{2} \Rightarrow \angle B A_{1} C=\frac{\pi}{2}
\end{array}
$$

and
$$
\frac{A_{1} B}{B C}=\frac{1}{2} \Rightarrow \angle A_{1} B C=\frac{\pi}{3}
$$

Also, $\angle A_{1} B A=\frac{\pi}{2}, \angle A B C=\frac{\pi}{2}$, so the dihedral angle $A_{1}-B A-C$ is $\frac{\pi}{3}$.
And $\sin \angle B_{1} B A=\sin \angle A_{1} A B=\frac{A_{1} B}{A A_{1}}=\frac{\sqrt{3}}{3}$,
so $\cos \angle B_{1} B C=\cos \angle B_{1} B A \cos \angle A B C+\sin \angle B_{1} B A \sin \angle A B C \cos \left(A_{1}\right.$ $\left.-B A-C) \cdot \cos \angle B_{1} B C\right)=\frac{\sqrt{3}}{3} \cdot 1 \cdot \frac{1}{2}$ (since $\angle A B C=\frac{\pi}{2}$ ) $=\frac{\sqrt{3}}{6}$
Therefore, $\angle B_{1} C_{1} C=\angle B_{1} B C=\arccos \frac{\sqrt{3}}{6}$.