Square $ABCD$ has side length $5$ and arc $BD$ with center $A$. $E$ is the midpoint of $AB$ and $CE$ intersects arc $BD$ at $F$. $G$ is placed onto $BC$ such that $FG$ is perpendicular to $BC$. What is the length of $FG$?
1. **Determine the length of \( AE \) and \( BE \):**
   Since \( E \) is the midpoint of \( \overline{AB} \), we have:
   \[
   AE = BE = \frac{5}{2}
   \]

2. **Calculate \( CE \) using the Pythagorean Theorem:**
   \[
   CE^2 = BC^2 + BE^2
   \]
   Given \( BC = 5 \) and \( BE = \frac{5}{2} \):
   \[
   CE^2 = 5^2 + \left( \frac{5}{2} \right)^2 = 25 + \frac{25}{4} = \frac{100}{4} + \frac{25}{4} = \frac{125}{4}
   \]
   Therefore:
   \[
   CE = \sqrt{\frac{125}{4}} = \frac{5\sqrt{5}}{2}
   \]

3. **Use the Power of a Point theorem:**
   Since \( AB \) is a chord and \( E \) is a point on \( AB \):
   \[
   \left( \frac{5}{2} \right)^2 = \frac{5\sqrt{5}}{2} \cdot EF
   \]
   Solving for \( EF \):
   \[
   \frac{25}{4} = \frac{5\sqrt{5}}{2} \cdot EF
   \]
   \[
   EF = \frac{25}{4} \cdot \frac{2}{5\sqrt{5}} = \frac{25 \cdot 2}{4 \cdot 5\sqrt{5}} = \frac{50}{20\sqrt{5}} = \frac{5}{2\sqrt{5}} = \frac{\sqrt{5}}{2}
   \]

4. **Calculate \( CF \):**
   Since \( CE = \frac{5\sqrt{5}}{2} \) and \( EF = \frac{\sqrt{5}}{2} \):
   \[
   CF = CE - EF = \frac{5\sqrt{5}}{2} - \frac{\sqrt{5}}{2} = \frac{4\sqrt{5}}{2} = 2\sqrt{5}
   \]

5. **Use similarity of triangles \( \triangle CBE \) and \( \triangle CGF \):**
   By AA similarity, we have:
   \[
   \frac{BE}{CE} = \frac{GF}{CF}
   \]
   Substituting the known values:
   \[
   \frac{\frac{5}{2}}{\frac{5\sqrt{5}}{2}} = \frac{FG}{2\sqrt{5}}
   \]
   Simplifying:
   \[
   \frac{5}{2} \cdot \frac{2}{5\sqrt{5}} = \frac{FG}{2\sqrt{5}}
   \]
   \[
   \frac{1}{\sqrt{5}} = \frac{FG}{2\sqrt{5}}
   \]
   Solving for \( FG \):
   \[
   FG = 2\sqrt{5} \cdot \frac{1}{\sqrt{5}} = 2
   \]

The final answer is \( \boxed{2} \).