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 $m+n$?
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$
Since $\angle{AFB}=\angle{ADB}=90^{\circ}$, quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$, so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE \sim \triangle ACD$. We can easily find $AD=12$, $BD = 5$, and $DC=9$ using Pythagorean triples. 
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is $\tfrac{12}{15} = \tfrac{4}{5}$, and the ratio of the shorter leg to the hypotenuse is $\tfrac{9}{15} = \tfrac{3}{5}$. It follows that $AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)$. 
Let $x=DF$. By [Ptolemy's Theorem](https://artofproblemsolving.com/wiki/index.php/Ptolemy%27s_Theorem), we have \[13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.\] Dividing by $13$ we get $x+4=7.2\implies x=\frac{16}{5}$ so our answer is $\boxed{\textbf{(B) }21}$.
~Edits by BakedPotato66