10. As shown in Figure 1, in $\triangle A B C$, it is known that
$\cos C=\frac{2 \sqrt{5}}{5}$,
$\overrightarrow{A H} \cdot \overrightarrow{B C}=0$,
$\overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$.
Figure 1
Then the eccentricity of the hyperbola passing through point $C$ and having $A, H$ as its two foci is . $\qquad$
10. $2+\sqrt{5}$.

Take the midpoint $D$ of side $AB$. Then $\overrightarrow{CA}+\overrightarrow{CB}=2\overrightarrow{AD}$.
$$
\begin{array}{l}
\text{By } \overrightarrow{AB} \cdot (\overrightarrow{CA}+\overrightarrow{CB})=0 \\
\Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AD}=0 \Rightarrow AB \perp AD.
\end{array}
$$

Thus, the altitude from $C$ to the base $AB$ of $\triangle ABC$ coincides with the median.
Hence, $\triangle ABC$ is an isosceles triangle, i.e., $AB=AC$.
By $\overrightarrow{AH} \cdot \overrightarrow{BC}=0$, we know $AH \perp BC$.
Given $\cos C=\frac{2\sqrt{5}}{5}$, we have
$\sin C=\frac{\sqrt{5}}{5}, \tan C=\frac{1}{2}$.
In the right triangle $\triangle ACH$, let $CH=2$.
Thus, $AH=1$,
$$
BC=AC=\sqrt{AH^2+CH^2}=\sqrt{5} \text{.}
$$

Therefore, the eccentricity of the hyperbola with foci at $A$ and $H$ is
$$
e=\frac{AH}{AC-CH}=\frac{1}{\sqrt{5}-2}=\sqrt{5}+2 \text{.}
$$