Example 6 In the complete quadrilateral CFBEGA, the line containing diagonal $CE$ intersects the circumcircle of $\triangle ABC$ at point $D$. The circle passing through point $D$ and tangent to $FG$ at point $E$ intersects $AB$ at point $M$. Given $\frac{AM}{AB}=t$. Find $\frac{GE}{EF}$ (expressed in terms of $t$).
Solution: As shown in Figure 12, connect $A D$, $M D$, and $B D$. We have
$$
\begin{array}{l}
\angle D M A \\
=\angle D E G \\
=\angle F E C, \\
\angle F G E \\
=\angle M A D,
\end{array}
$$

Thus, $\triangle E F C \backsim \triangle M D A$.
Therefore, $\frac{E F}{M D}=\frac{C E}{A M}$, which means
$$
E F \cdot A M=M D \cdot C E \text {. }
$$

Also, since $\angle F E C=\angle D M E$, we know
$$
\begin{array}{l}
\angle G E C=\angle D M B, \\
\angle E C G=\angle D B M .
\end{array}
$$

Thus, $\triangle G E C \backsim \triangle D M B$.
Therefore, $\frac{E G}{M D}=\frac{C E}{B M}$, which means
$$
E G \cdot B M=M D \cdot C E \text {. }
$$

Hence, $E F \cdot M A=E G \cdot B M$.
Therefore, $\frac{G E}{E F}=\frac{A M}{M B}=\frac{A M}{A B-A M}$
$$
=\frac{t A B}{A B-t A B}=\frac{t}{1-t} .
$$