9. In $\triangle A B C$, $\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=0, \overrightarrow{G A} \cdot \overrightarrow{G B}=0$. $\frac{(\tan A+\tan B) \tan C}{\tan A \tan B}=$ $\qquad$ .
Solve:
As shown in the figure, by the median formula we get
$$
\begin{aligned}
& A D^{2}=\frac{2 b^{2}+2 c^{2}-a^{2}}{4}, B E^{2}=\frac{2 a^{2}+2 c^{2}-b^{2}}{4}, \\
& \text { In } \triangle A B G, A G^{2}+B G^{2}=A B^{2} \\
& \frac{2 b^{2}+2 c^{2}-a^{2}}{9}+\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=c^{2} \\
\Rightarrow & a^{2}+b^{2}=5 c^{2} \cdot \text { And } \frac{(\tan A+\tan B) \tan C}{\tan A \tan B}=\frac{\sin ^{2} C}{\sin A \sin B \cos C} \\
= & \frac{c^{2}}{a b} \cdot \frac{2 a b}{a^{2}+b^{2}-c^{2}}=\frac{1}{2} .
\end{aligned}
$$