9. As shown in the figure, in $\triangle A B C$, $\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}, \overrightarrow{A H} \cdot \overrightarrow{B C}=0, \overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$, then the eccentricity of the hyperbola passing through point $C$, with $A, H$ as the two foci is $\qquad$ .
Given, $\cos C=2 \cos ^{2} \frac{C}{2}-1=\frac{3}{5}, \sin C=\frac{4}{5}$, let's assume $A C=B C=5$, then $A H=4, C H=3$. Therefore, for the hyperbola, $2 a=2,2 c=4 \Rightarrow e=\frac{c}{a}=2$.