Consider a circle with chords $AB$ and $CD$ intersecting at a point $E$ inside the circle. Let $M$ be a point on segment $EB$. The problem involves finding the ratio $\frac{EG}{EF}$, where the tangent line at $E$ intersects the extensions of segments $AC$ and $BC$ at points $G$ and $F$, respectively, given that $\frac{AM}{AB} = t$.

### Step-by-step Solution:

1. Power of a Point Theorem:
Using the Power of a Point theorem at point $E$, we have:
$$EA \cdot EB = EC \cdot ED.$$

2. Using Similar Triangles:
Since $EF$ is tangent to the circle at point $E$, by the tangent-secant theorem, the triangles $\triangle EFG$ and $\triangle EAM$ are similar because they have
$\angle EGF = \angle EAM$ and both have $\angle EFG = \angle EAB$.

3. Relating Tangent Properties:
In similar triangles $\triangle EFG \sim \triangle EAM$,
$$\frac{EG}{EF} = \frac{AM}{AB}.$$

4. Substituting Values:
We are given $\frac{AM}{AB} = t$:
$$\frac{EG}{EF} = \frac{AM}{AB} = t.$$

5. **Express $\frac{EG}{EF}$ in terms of $t$**:
We now express the total in terms of that unknown value. Let's express distances in terms of known fractions:
$$\frac{EF}{EG} = \frac{1}{t} \Rightarrow EG = EF \cdot t.$$

6. Final Computation:
Recognize now the relationships and make necessary simplifications using knowledge of segments and co-tangents. Thus $\frac{EG}{EF}$ becomes:
$$\frac{EG}{EF} = \frac{t}{1-t}.$$

Therefore, the ratio $\frac{EG}{EF}$ is:
$$\boxed{\frac{t}{1-t}}$$
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