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

Then the eccentricity of the hyperbola passing through point $C$ and having $A$ and $H$ as its foci is . $\qquad$
9. 2 .

Given $\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow(\overrightarrow{C B}-\overrightarrow{C A}) \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$
$\Rightarrow A C=B C$.
From $\overrightarrow{A H} \cdot \overrightarrow{B C}=0 \Rightarrow A H \perp B C$.
Since $\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}$, therefore,
$\sin \frac{C}{2}=\frac{\sqrt{5}}{5}, \tan \frac{C}{2}=\frac{1}{2}$.
Then $\tan C=\frac{2 \tan \frac{C}{2}}{1-\tan ^{2} \frac{C}{2}}=\frac{4}{3}$.
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
$$
e=\frac{A H}{A C-C H}=2 .
$$