Consider a regular pentagon $ABCDE$, and let the intersection of diagonals $\overline{CA}$ and $\overline{EB}$ be $F$. Find $\angle AFB$.

[i]Proposed by Justin Chen[/i]
1. **Identify the properties of a regular pentagon:**
   - A regular pentagon has all sides of equal length and all interior angles equal.
   - Each interior angle of a regular pentagon is given by:
     \[
     \frac{(n-2) \times 180^\circ}{n} = \frac{(5-2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = 108^\circ
     \]

2. **Consider the diagonals of the pentagon:**
   - In a regular pentagon, the diagonals intersect at points that divide the angles into smaller angles.
   - Specifically, the diagonals $\overline{CA}$ and $\overline{EB}$ intersect at point $F$.

3. **Analyze the triangle $\triangle ABF$:**
   - Since $ABCDE$ is a regular pentagon, the diagonals $\overline{CA}$ and $\overline{EB}$ intersect at $F$ such that $\angle AFB$ is formed.
   - The diagonals of a regular pentagon intersect in such a way that they divide the angles at the vertices into smaller angles of $36^\circ$ and $72^\circ$.

4. **Determine the angles in $\triangle ABF$:**
   - In $\triangle ABF$, $\angle BAF$ and $\angle ABF$ are each $36^\circ$ because they are part of the angles formed by the intersection of the diagonals.
   - Therefore, $\angle BAF = \angle ABF = 36^\circ$.

5. **Calculate $\angle AFB$:**
   - Using the fact that the sum of the angles in a triangle is $180^\circ$, we can find $\angle AFB$:
     \[
     \angle AFB = 180^\circ - \angle BAF - \angle ABF = 180^\circ - 36^\circ - 36^\circ = 108^\circ
     \]

Thus, the measure of $\angle AFB$ is $108^\circ$.

The final answer is $\boxed{108^\circ}$