9. As shown in Figure 1, in $\triangle A B C$, $\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}$, $\overrightarrow{A H} \cdot \overrightarrow{B C}=0$,
$$
\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0 \text {. }
$$

Then the eccentricity of the hyperbola passing through point $C$ and having $A$ and $H$ as its foci is $\qquad$
9. 2 .
$$
\begin{array}{l}
\text { Given } \overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0 \\
\Rightarrow(\overrightarrow{C B}-\overrightarrow{C A}) \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0 \\
\Rightarrow A C=B C .
\end{array}
$$
$$
\text { Given } \overrightarrow{A H} \cdot \overrightarrow{B C}=0 \Rightarrow A H \perp B C \text {. }
$$

Since $\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}$, then,
$$
\sin \frac{C}{2}=\frac{\sqrt{5}}{5}, \tan \frac{C}{2}=\frac{1}{2} \text {. }
$$

Thus, $\tan C=\frac{2 \tan \frac{C}{2}}{1-\tan ^{2} \frac{C}{2}}=\frac{4}{3}$.
In the right triangle $\triangle A C H$, let's assume $C H=3$.
Then $A H=4, A C=B C=\sqrt{A H^{2}+C H^{2}}=5$. Therefore, the eccentricity of the hyperbola with foci at $A$ and $H$ is
$$
e=\frac{A H}{A C-C H}=2 \text {. }
$$