Example 2 As shown in Figure 2, given that $D$ is the intersection of the tangents to the circumcircle $\odot O$ of $\triangle A B C$ at points $A$ and $B$, the circumcircle of $\triangle A B D$ intersects line $A C$ and segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. If $\frac{B C}{B F}=2$, find $\frac{B G}{G E}$.
(2010-2011, Hungarian Mathematical Olympiad)
Let $C D$ intersect $A B$ at point $M$.
If the center $O$ is not on the line segment $B C$, from $\frac{B C}{B F}=2$, we know that $F$ is the midpoint of $B C$.
Then, by the perpendicular diameter theorem,
$$
F O \perp B C \Rightarrow \angle O F B=90^{\circ} \text {. }
$$

Since $D A$ and $D B$ are tangent to $\odot O$, we have
$$
\angle O A D=\angle O B D=90^{\circ} \text {. }
$$

Thus, $O, A, B, D$ are concyclic.
Since $F, A, B, D$ are concyclic, then $F, O, A, B, D$ are concyclic. Therefore,
$$
\angle O A B=\angle O F B=90^{\circ} \text {. }
$$

This contradicts $\angle O A D=90^{\circ}$.
Therefore, point $F$ coincides with $O$.
Next, consider the ratio of some segments
$$
\begin{aligned}
\frac{A M}{M B} & =\frac{S_{\triangle A C D}}{S_{\triangle B C D}}=\frac{\frac{1}{2} A C \cdot A D \sin \angle C A D}{\frac{1}{2} B C \cdot B D \sin \angle C B D} \\
= & \frac{A C}{B C} \cdot \frac{A D}{B D} \cdot \frac{\sin \left(90^{\circ}+\angle C A F\right)}{1} \\
= & \frac{A C}{B C} \cdot \frac{A D}{B D} \cos \angle C A F .
\end{aligned}
$$

Since $\angle C A F=\angle A C F=\angle B C A$, by the tangent length theorem, we know $A D=B D$.
Thus, $\frac{A M}{B M}=\frac{A C}{B C} \cos \angle B C A=\left(\frac{A C}{B C}\right)^{2}$.
Applying Menelaus' theorem to $\triangle B A E$ and the transversal $G M C$, we get
$$
\frac{B G}{G E} \cdot \frac{E C}{C A} \cdot \frac{A M}{M B}=1 \text {. }
$$

Therefore, $\frac{B G}{G E}=\frac{C A}{C E} \cdot \frac{M B}{M A}=\left(\frac{B C}{A C}\right)^{2} \frac{A C}{C E}=\frac{B C^{2}}{A C \cdot C E}$.
By the secant theorem and $F$ being the midpoint of $B C$, we have
$$
A C \cdot C E=C F \cdot C B=\frac{1}{2} B C^{2} \text {. }
$$

Therefore, $\frac{B G}{G E}=2$.