G59 (31-1, India) Let two chords $AB, CD$ of a circle intersect at a point $E$ inside the circle. Take a point $M$ inside the line segment $EB$, and then draw a circle through points $D, E, M$. Draw a tangent line to this circle at $E$, which intersects the lines $BC, AC$ at points $F, G$ respectively. If it is known that $\frac{AM}{AB}=t$, try to express $\frac{EG}{EF}$ in terms of $t$.

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The translation is provided as requested, maintaining the original formatting and structure.
Connect $A D, M D$ and $B D$.
It is easy to see that
$$
\begin{array}{c}
\angle C E F=\angle D E G=\angle E M D, \\
\angle E C F=\angle M A D,
\end{array}
$$

Thus,
$$
\triangle C E F \backsim \triangle A D M,
$$

Therefore,
$$
C E \cdot M D=A M \cdot E F.
$$

Moreover, we also have
$$
\begin{array}{c}
\angle E C G=\angle M B D, \\
\angle C E G=\angle 180^{\circ}-\angle G E D \\
=180^{\circ}-\angle E M D \\
=\angle B M D,
\end{array}
$$

Hence, $\triangle C G E \sim \triangle B D M$, which implies $G E \cdot M B=C E \cdot M D$.
Thus, we have $G E \cdot M B=A M \cdot E F$, so
$$
\frac{E G}{E F}=\frac{A M}{M B}=\frac{t \cdot A B}{(1-t) \cdot A B}=\frac{t}{1-t}.
$$