23. 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, satisfying $A D \perp B C, D E \perp A C, A F \perp B F$, the length of segment $D F$ is a reduced fraction $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$. Then $m+n=(\quad)$.
(A) 18
(B) 21
(C) 24
(D) 27
(E) 30
23. B.

As shown in Figure 5.
Let $p$ be the semi-perimeter of $\triangle ABC$. Then
$$
p=\frac{a+b+c}{2}=\frac{13+14+15}{2}=21.
$$

Thus, $S_{\triangle ABC}=\sqrt{p(p-a)(p-b)(p-c)}$
$$
\begin{array}{l}
=\sqrt{21 \times 8 \times 7 \times 6}=84 \\
\Rightarrow AD=\frac{2 S_{\triangle ABC}}{BC}=\frac{2 \times 84}{14}=12.
\end{array}
$$

Since $AD \perp BC$, we have
$$
BD=\sqrt{AB^2-AD^2}=\sqrt{13^2-12^2}=5.
$$

Given $BC=14$, then $DC=9$.
Since $DE \perp AC$, we have
$$
\begin{array}{l}
DE=\frac{AD \cdot DC}{AC}=\frac{12 \times 9}{15}=\frac{36}{5}, \\
AE=\frac{AD^2}{AC}=\frac{12^2}{15}=\frac{48}{5}.
\end{array}
$$

Since $\angle AFB=\angle ADB=90^\circ$, points $A, B, D, F$ are concyclic.
Therefore, $\angle AFE=\angle ABC$.
Thus, $\cot \angle AFE=\cot \angle ABC=\frac{5}{12}$.
In $\triangle AEF$,
$$
\begin{array}{l}
EF=AE \cot \angle AFE=\frac{48}{5} \times \frac{5}{12}=4 \\
\Rightarrow DF=DE-EF=\frac{36}{5}-4=\frac{16}{5} \\
\Rightarrow m=16, n=5 \\
\Rightarrow m+n=21.
\end{array}
$$