Right $ \triangle ABC$ has $ AB \equal{} 3$, $ BC \equal{} 4$, and $ AC \equal{} 5$. Square $ XYZW$ is inscribed in $ \triangle ABC$ with $ X$ and $ Y$ on $ \overline{AC}$, $ W$ on $ \overline{AB}$, and $ Z$ on $ \overline{BC}$. What is the side length of the square?

[asy]size(200);defaultpen(fontsize(10pt)+linewidth(.8pt));
real s = (60/37);
pair A = (0,0);
pair C = (5,0);
pair B = dir(60)*3;
pair W = waypoint(B--A,(1/3));
pair X = foot(W,A,C);
pair Y = (X.x + s, X.y);
pair Z = (W.x + s, W.y);

label("$A$",A,SW);
label("$B$",B,NW);
label("$C$",C,SE);
label("$W$",W,NW);
label("$X$",X,S);
label("$Y$",Y,S);
label("$Z$",Z,NE);

draw(A--B--C--cycle);
draw(X--W--Z--Y);[/asy]

$ \textbf{(A)}\ \frac {3}{2}\qquad \textbf{(B)}\ \frac {60}{37}\qquad \textbf{(C)}\ \frac {12}{7}\qquad \textbf{(D)}\ \frac {23}{13}\qquad \textbf{(E)}\ 2$
1. **Identify the right triangle and its dimensions:**
   Given a right triangle \( \triangle ABC \) with \( AB = 3 \), \( BC = 4 \), and \( AC = 5 \). This is a classic 3-4-5 right triangle.

2. **Position the square inside the triangle:**
   The square \( XYZW \) is inscribed in \( \triangle ABC \) such that \( X \) and \( Y \) are on \( \overline{AC} \), \( W \) is on \( \overline{AB} \), and \( Z \) is on \( \overline{BC} \).

3. **Set up the problem with variables:**
   Let the side length of the square be \( s \). The coordinates of the vertices of the triangle are \( A = (0, 0) \), \( B = (0, 3) \), and \( C = (4, 0) \).

4. **Determine the coordinates of the square's vertices:**
   - \( W \) is on \( \overline{AB} \), so \( W = (0, s) \).
   - \( Z \) is on \( \overline{BC} \), so \( Z = (s, 3 - \frac{3s}{4}) \).
   - \( X \) and \( Y \) are on \( \overline{AC} \), so \( X = (s, 0) \) and \( Y = (4 - \frac{4s}{5}, s) \).

5. **Use similarity of triangles:**
   The small triangle formed by the square and the larger triangle are similar. Specifically, \( \triangle WAX \sim \triangle ABC \).

6. **Set up the similarity ratio:**
   The height of \( \triangle ABC \) from \( B \) to \( AC \) is \( \frac{12}{5} \) (since the area of \( \triangle ABC \) is \( \frac{1}{2} \times 3 \times 4 = 6 \), and the height from \( B \) to \( AC \) is \( \frac{2 \times 6}{5} = \frac{12}{5} \)).

7. **Express the height in terms of \( s \):**
   The height of the triangle formed by the square is \( \frac{12}{5} - s \).

8. **Set up the equation using the similarity ratio:**
   \[
   \frac{s}{\frac{12}{5}} = \frac{5 - s}{5}
   \]
   Simplify the equation:
   \[
   \frac{5s}{12} = \frac{5 - s}{5}
   \]
   Cross-multiply to solve for \( s \):
   \[
   25s = 60 - 12s
   \]
   Combine like terms:
   \[
   37s = 60
   \]
   Solve for \( s \):
   \[
   s = \frac{60}{37}
   \]

The final answer is \( \boxed{\frac{60}{37}} \).