The triangle ABC has sides AB = 137, AC = 241, and BC =200. There is a point D, on BC, such that both incircles of triangles ABD and ACD touch AD at the same point E. Determine the length of CD.
[asy]

pair A = (2,6);
pair B = (0,0);
pair C = (10,0);
pair D = (3.5,0) ;
pair E = (3.1,2);
draw(A--B);
draw(B--C);
draw(C--A);
draw (A--D);
dot ((3.1,1.7));
label  ("E", E, dir(45));
label ("A", A, dir(45));
label ("B", B, dir(45));
label ("C", C, dir(45));
label ("D", D, dir(45));
draw(circle((1.8,1.3),1.3)); 
draw(circle((4.9,1.7),1.75)); 
[/asy]
1. **Define Variables and Tangent Points:**
   Let \( M \) be the tangent point on \( AB \), \( N \) be the tangent point on \( BD \), \( O \) be the tangent point on \( DC \), and \( P \) be the tangent point on \( AC \). Let \( E \) be the common tangent point on \( AD \) for both incircles of triangles \( ABD \) and \( ACD \).

2. **Assign Variables to Tangent Segments:**
   - Let \( AM = AE = AP = w \)
   - Let \( BM = BN = x \)
   - Let \( ND = DE = DO = y \)
   - Let \( CO = CP = z \)

3. **Set Up Equations Based on Tangent Properties:**
   Using the properties of tangents from a point to a circle, we can write the following equations:
   \[
   w + x = 137 \quad \text{(1)}
   \]
   \[
   w + z = 241 \quad \text{(2)}
   \]
   \[
   x + 2y + z = 200 \quad \text{(3)}
   \]
   \[
   w + x + y + z = 289 \quad \text{(4)}
   \]

4. **Solve the System of Equations:**
   - From equation (1): \( w = 137 - x \)
   - From equation (2): \( w = 241 - z \)
   - Equate the two expressions for \( w \):
     \[
     137 - x = 241 - z \implies z - x = 104 \quad \text{(5)}
     \]
   - Substitute \( w = 137 - x \) into equation (4):
     \[
     137 - x + x + y + z = 289 \implies 137 + y + z = 289 \implies y + z = 152 \quad \text{(6)}
     \]
   - Substitute \( y + z = 152 \) into equation (3):
     \[
     x + 2y + z = 200 \implies x + 2y + 152 - y = 200 \implies x + y + 152 = 200 \implies x + y = 48 \quad \text{(7)}
     \]

5. **Find \( y \) and \( z \):**
   - From equation (6): \( y + z = 152 \)
   - From equation (7): \( x + y = 48 \)
   - Substitute \( y = 48 - x \) into \( y + z = 152 \):
     \[
     (48 - x) + z = 152 \implies z = 152 - 48 + x \implies z = 104 + x
     \]
   - From equation (5): \( z - x = 104 \)
   - Substitute \( z = 104 + x \):
     \[
     104 + x - x = 104 \quad \text{(True)}
     \]

6. **Calculate \( CD \):**
   - From equation (3): \( x + 2y + z = 200 \)
   - Substitute \( y = 48 - x \) and \( z = 104 + x \):
     \[
     x + 2(48 - x) + 104 + x = 200 \implies x + 96 - 2x + 104 + x = 200 \implies 200 = 200 \quad \text{(True)}
     \]
   - Therefore, \( CD = z = 104 + x \).

7. **Determine \( x \):**
   - From equation (7): \( x + y = 48 \)
   - Substitute \( y = 48 - x \):
     \[
     x + (48 - x) = 48 \quad \text{(True)}
     \]

8. **Final Calculation:**
   - Since \( y = 48 - x \) and \( z = 104 + x \), we need to find \( CD \):
     \[
     CD = 200 - BD = 200 - (x + y) = 200 - 48 = 152
     \]

The final answer is \(\boxed{152}\)