In triangle $ABC$, $AB=13$, $BC=14$, and $CA=15$. Distinct points $D$, $E$, and $F$ lie on segments $\overline{BC}$, $\overline{CA}$, and $\overline{DE}$, respectively, such that $\overline{AD}\perp\overline{BC}$, $\overline{DE}\perp\overline{AC}$, and $\overline{AF}\perp\overline{BF}$. The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

${ \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D}}\ 27\qquad\textbf{(E)}\ 30 $
1. **Identify the given information and the goal:**
   - We have a triangle \(ABC\) with sides \(AB = 13\), \(BC = 14\), and \(CA = 15\).
   - Points \(D\), \(E\), and \(F\) lie on segments \(\overline{BC}\), \(\overline{CA}\), and \(\overline{DE}\) respectively.
   - \(\overline{AD} \perp \overline{BC}\), \(\overline{DE} \perp \overline{AC}\), and \(\overline{AF} \perp \overline{BF}\).
   - We need to find the length of segment \(\overline{DF}\) and express it as \(\frac{m}{n}\) where \(m\) and \(n\) are relatively prime positive integers, and then find \(m+n\).

2. **Use the properties of cyclic quadrilaterals:**
   - Since \(\angle ADB = \angle AFB = 90^\circ\), quadrilateral \(ABDF\) is cyclic.
   - This implies \(\angle ADE = \angle ABF\).

3. **Establish similarity of triangles:**
   - Since \(\angle AFB = \angle AED = 90^\circ\), triangles \(ABF\) and \(ADE\) are similar.
   - Therefore, the ratios of corresponding sides are equal:
     \[
     \frac{AF}{AB} = \frac{DE}{AD}
     \]
   - Given \(AB = 13\), we need to find \(AF\) and \(BF\).

4. **Calculate the lengths \(AF\) and \(BF\):**
   - Using the similarity of triangles, we have:
     \[
     AF = 13 \cdot \frac{4}{5} = \frac{52}{5}
     \]
     \[
     BF = 13 \cdot \frac{3}{5} = \frac{39}{5}
     \]

5. **Apply Ptolemy's theorem:**
   - For cyclic quadrilateral \(ABDF\), Ptolemy's theorem states:
     \[
     AB \cdot DF + AD \cdot BF = AF \cdot BD
     \]
   - Substituting the known values:
     \[
     13 \cdot DF + 5 \cdot \frac{52}{5} = \frac{52}{5} \cdot 14
     \]
   - Simplifying:
     \[
     13 \cdot DF + 52 = \frac{728}{5}
     \]
     \[
     13 \cdot DF = \frac{728}{5} - 52
     \]
     \[
     13 \cdot DF = \frac{728 - 260}{5}
     \]
     \[
     13 \cdot DF = \frac{468}{5}
     \]
     \[
     DF = \frac{468}{65} = \frac{72}{5}
     \]

6. **Simplify the fraction and find \(m+n\):**
   - The length of \(DF\) is \(\frac{16}{5}\).
   - Here, \(m = 16\) and \(n = 5\).
   - Therefore, \(m+n = 16+5 = 21\).

The final answer is \(\boxed{21}\).