9. As shown in Figure 2, in $\triangle A B C$, $A B=A C$, $A D \perp B C$, $C G / / A B$, $B G$ intersects $A D$ and $A C$ at points $E$ and $F$ respectively. If $\frac{E F}{B E}=\frac{a}{b}$, then $\frac{G E}{B E}$ equals $\qquad$
Two $9 \cdot \frac{b}{a}$.
As shown in Figure 6, connect $C E$ and extend it to intersect $A B$ at point $H$. Clearly, $E H=E F, B E=E C$. Since $C G / / A B$, we know $\triangle E H B \backsim \triangle E C G$. Therefore, we have
$$
\begin{array}{l}
\frac{G E}{E B}=\frac{E C}{E H} \\
=\frac{B E}{E F}=\frac{b}{a} .
\end{array}
$$