Let's analyze the given triangle $\triangle ABC$ with the following details:

1. $BC = 60$
2. $AD = 20$
3. Points $E$ and $G$ lie on a one-quarter circle centered at $F$.

Since $E$ and $G$ are on the quarter-circle, $F$ serves as the center of this arc, and lines $DF$ and $EF$ are radii of this circle segment. Additionally, the one-quarter circle implies a right angle at $D$ and $E$, so $\angle DEF = 90^\circ$.

Steps to Solve for $EF$

1. Recognize Right Triangle Properties: Since $\angle DEF = 90^\circ$ and $AD = 20$, the radius of the quarter-circle is $DF = EF$. So, we need to find the radius $EF$ in terms of the geometry of $\triangle ABC$ and the quarter-circle.

2. Use Right Triangle Proportions: Given the setup, $\triangle DEF$ forms a right triangle where $DF = EF$, and both are half the length from $D$ to $A$ (since $AD = 20$ and forms a perpendicular distance from $B$ and $C$).

Therefore: $EF = \frac{AD}{2} = \frac{20}{2} = 10$

Conclusion

The length of $EF$ is 10 units.

Would you like further details on any of the steps?

Here are some follow-up questions related to the concepts used in this problem:

1. How do right triangle properties apply in a one-quarter circle configuration?
2. What additional properties can be derived from the lengths in $\triangle ABC$?
3. How would the solution change if $\angle BAC$ was not a right angle?
4. How does the concept of radius affect the solution when dealing with quarter circles?
5. What other methods could be used to verify the length of $EF$ in this configuration?

Tip: When dealing with quarter circles in geometry, focus on symmetry and radius properties as they often simplify calculations.