Let the vertices $A$ and $C$ of a right triangle $ABC$ be on the arc with center $B$ and radius $20$. A semicircle with diameter $[AB]$ is drawn to the inner region of the arc. The tangent from $C$ to the semicircle touches the semicircle at $D$ other than $B$. Let $CD$ intersect the arc at $F$. What is $|FD|$?

$ 
\textbf{(A)}\ 1
\qquad\textbf{(B)}\ \frac 52
\qquad\textbf{(C)}\ 3
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ 5
$
1. **Identify the given elements and their relationships:**
   - \( A \) and \( C \) are on an arc with center \( B \) and radius \( 20 \).
   - A semicircle with diameter \( AB \) is drawn inside the arc.
   - \( C \) is tangent to the semicircle at \( D \).
   - \( CD \) intersects the arc at \( F \).

2. **Define the center of the semicircle:**
   Let \( O \) be the center of the semicircle. Since the diameter of the semicircle is \( AB \), \( O \) is the midpoint of \( AB \).

3. **Establish the properties of the right kite \( BCDO \):**
   - \( B \) is the center of the arc, so \( BC = 20 \).
   - \( O \) is the midpoint of \( AB \), so \( BO = 10 \).
   - \( D \) is the point of tangency, so \( OD \) is perpendicular to \( CD \).

4. **Calculate \( BD \):**
   Since \( BCDO \) forms a right kite, we can use the Pythagorean theorem in \( \triangle BOD \):
   \[
   BD = \sqrt{BO^2 + OD^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}
   \]

5. **Use Stewart's Theorem in \( \triangle BCF \):**
   Stewart's Theorem states:
   \[
   BC^2 \cdot AF + CA^2 \cdot BF = AB^2 \cdot CF + CF \cdot AF \cdot BF
   \]
   Here, \( BC = 20 \), \( CA = 20 \), and \( AB = 20 \). Since \( \triangle BCF \) is isosceles, \( AF = BF \).

6. **Simplify using the given values:**
   \[
   20^2 \cdot AF + 20^2 \cdot BF = 20^2 \cdot CF + CF \cdot AF \cdot BF
   \]
   Since \( AF = BF \), let \( x = AF = BF \):
   \[
   400x + 400x = 400 \cdot CF + CF \cdot x \cdot x
   \]
   Simplifying, we get:
   \[
   800x = 400 \cdot CF + CF \cdot x^2
   \]

7. **Solve for \( x \):**
   Since \( CF = 20 \) and \( x = 4 \):
   \[
   20^2 - 20x = (10\sqrt{2})^2 \Rightarrow 400 - 20x = 200 \Rightarrow 20x = 200 \Rightarrow x = 4
   \]

8. **Conclusion:**
   The length \( |FD| \) is \( 4 \).

The final answer is \( \boxed{4} \).