A semicircle has diameter $\overline{AD}$ with $AD = 30$. Points $B$ and $C$ lie on $\overline{AD}$, and points $E$ and $F$ lie on the arc of the semicircle. The two right triangles $\vartriangle BCF$ and $\vartriangle CDE$ are congruent. The area of $\vartriangle  BCF$ is $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m + n$.
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1. **Identify the given information and setup the problem:**
   - The diameter \( \overline{AD} \) of the semicircle is \( AD = 30 \).
   - Points \( B \) and \( C \) lie on \( \overline{AD} \).
   - Points \( E \) and \( F \) lie on the arc of the semicircle.
   - The right triangles \( \triangle BCF \) and \( \triangle CDE \) are congruent.
   - We need to find the area of \( \triangle BCF \) in the form \( m\sqrt{n} \) and then find \( m + n \).

2. **Establish congruence and symmetry:**
   - Since \( \triangle BCF \) and \( \triangle CDE \) are congruent, their corresponding sides and angles are equal.
   - Reflecting \( \triangle CDE \) over the vertical radius of the semicircle, we get \( \triangle AFB \). Thus, \( \triangle BCF \cong \triangle CDE \cong \triangle AFB \).

3. **Determine the positions of points \( B \) and \( C \):**
   - Since \( \triangle BCF \) and \( \triangle CDE \) are congruent and symmetric about the center of the semicircle, \( B \) and \( C \) must divide \( \overline{AD} \) into three equal segments.
   - Therefore, \( AB = BC = CD = \frac{30}{3} = 10 \).

4. **Calculate the coordinates of points \( B \) and \( C \):**
   - Let the center of the semicircle be \( O \) at the midpoint of \( \overline{AD} \).
   - \( O \) is at \( (15, 0) \).
   - \( B \) is at \( (5, 0) \) and \( C \) is at \( (15, 0) \).

5. **Determine the length of \( BF \):**
   - Since \( F \) lies on the semicircle, its coordinates are \( (x, y) \) such that \( x^2 + y^2 = 15^2 = 225 \).
   - \( B \) is at \( (5, 0) \), so the distance \( BF \) is:
     \[
     BF = \sqrt{(x - 5)^2 + y^2}
     \]
   - Since \( F \) is on the semicircle and \( \triangle BCF \) is a right triangle, \( F \) must be directly above \( B \) on the semicircle. Thus, \( x = 5 \) and \( y = \sqrt{225 - 25} = \sqrt{200} = 10\sqrt{2} \).

6. **Calculate the area of \( \triangle BCF \):**
   - The base \( BC = 10 \) and the height from \( B \) to \( F \) is \( 10\sqrt{2} \).
   - The area of \( \triangle BCF \) is:
     \[
     \text{Area} = \frac{1}{2} \times BC \times BF = \frac{1}{2} \times 10 \times 10\sqrt{2} = 50\sqrt{2}
     \]

7. **Identify \( m \) and \( n \) and find \( m + n \):**
   - The area is \( 50\sqrt{2} \), so \( m = 50 \) and \( n = 2 \).
   - Therefore, \( m + n = 50 + 2 = 52 \).

The final answer is \( \boxed{52} \).