19. In $\triangle A B C$, it is known that $A B=13, B C=14, C A=$ 15, points $D, E, F$ are on sides $B C, C A, D E$ respectively, and $A D \perp$ $B C, D E \perp A C, A F \perp B F$. If the length of segment $D F$ is $\frac{m}{n}$ $\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$, then $m+n=(\quad)$.
(A) 18
(B) 21
(C) 24
(D) 27
(E) 30
19. B.

As shown in Figure 3, with $D$ as the origin, the line $BC$ as the $x$-axis, and the line $AD$ as the $y$-axis, we establish a Cartesian coordinate system $x D y$.
It is easy to know that $p_{\triangle ABC}=13+14+15=42$,
$S_{\triangle ABC}=\sqrt{21(21-13) \times(21-14) \times(21-15)}=84$.
Thus, $AD=\frac{2 S_{\triangle ABC}}{BC}=12$.
Therefore, the coordinates of point $A(0,12), B(-5,0), C(9,0)$.
Since $DE \perp AC$, the equation of line $DE$ is
$$
y=\frac{3}{4} x \text {. }
$$

Because $AF \perp BF$, point $F$ lies on the circumcircle with $AB$ as the diameter.
Note that the equation of the circle with $AB$ as the diameter is
$$
\left(x+\frac{5}{2}\right)^{2}+(y-6)^{2}=\left(\frac{13}{2}\right)^{2} \text {. }
$$

Substituting equation (1) into equation (2) yields
$$
\begin{array}{l}
x^{2}+5 x+\frac{25}{4}+\frac{9}{16} x^{2}-9 x+36=\frac{169}{4} \\
\Rightarrow \frac{25}{16} x^{2}-4 x=0 \Rightarrow x_{1}=0, x_{2}=\frac{64}{25} .
\end{array}
$$

Therefore, point $F\left(\frac{64}{25}, \frac{48}{25}\right)$.
Thus, $DF=\sqrt{\left(\frac{64}{25}\right)^{2}+\left(\frac{48}{25}\right)^{2}}=\frac{16}{5}$.
Hence, $m+n=21$.