Example 1 Let two chords $AB$ and $CD$ of a circle intersect at a point $E$ inside the circle. Take a point $M$ on the line segment $EB$ internally, then draw a circle through points $D$, $E$, and $M$. Draw a tangent to this circle at $E$ that intersects lines $BC$ and $AC$ at points $F$ and $G$ respectively. If $\frac{AM}{MB}=t$, try to express $\frac{EG}{EF}$ in terms of $t$.
Solution 1: As shown in Figure 12-1, draw line segments from $D$ to $A, B, M$. Clearly, $\angle C E F = \angle D E G = \angle E M D, \angle E C F = \angle M A D$. Therefore, $\triangle C E F \sim \triangle A M D$. Thus, $C E \cdot M D = A M \cdot E F$.

On the other hand, we also have $\angle E C G = \angle M B D$, so $\angle C G E = \angle C E F = \angle E M D - \angle M B D = \angle B D M$. Hence, $\triangle C G E \sim \triangle B D M$. Thus, $G E \cdot M B = C E \cdot M D$. Therefore, $G E \cdot M B = A M \cdot E F$.
Thus, $\frac{G E}{E F} = \frac{A M}{M B} = \frac{t(A B)}{(1-t) A B} = \frac{t}{1-t}$.

Solution 2: As shown in Figure 12-1, we have
$$
\begin{array}{l}
\angle A E G = \angle F E B = \angle E D M, \\
\angle F E C = \angle G E D = \angle D M E .
\end{array}
$$

Therefore, using the relationship between the sides, angles, and areas of triangles, we get
$$
\begin{array}{l}
\frac{S_{\triangle B E C}}{S_{\triangle F E C}} = 1 + \frac{S_{\triangle F E B}}{S_{\triangle F E C}} \cdot \frac{S_{\triangle D M E}}{S_{\triangle M D E}} \\
= 1 + \frac{B E \cdot M E}{D E \cdot C E} = 1 + \frac{B E \cdot M E}{B E \cdot A E} \\
= \frac{A M}{A E} ; \\
\frac{S_{\triangle A E C}}{S_{\triangle G E C}} = 1 - \frac{S_{\triangle G E A}}{S_{\triangle G E C}} \cdot \frac{S_{\triangle D M C}}{S_{\triangle M D E}} \\
\quad = 1 - \frac{A E \cdot M E}{D E \cdot C E} = 1 - \frac{A E \cdot M E}{A E \cdot B E} = \frac{B M}{B E} .
\end{array}
$$

Using the above results, we get
$$
\frac{G E}{E F} = \frac{S_{\triangle G E C}}{S_{\triangle F E C}} = \frac{A M}{B M} = \frac{t}{1-t} .
$$