2. As shown in the figure, from the obtuse vertex $A$ of the obtuse triangle $\triangle ABC$, draw the altitude $AD$. With the foot of the perpendicular $D$ as the center and $AD$ as the radius, draw a circle that intersects $AB$ and $AC$ at $M$ and $N$, respectively. If $AB=c, AM=m, AN=n$, then the length of side $AC$ is ( ).
(A) $m+n$
(B) $\frac{nc}{m}$
(C) $\frac{mc}{n}$
(D) $\left(\frac{n}{m}+\frac{m}{n}\right)c$
2. (C).

As shown in the figure, extend $A D$ to intersect $\odot D$ at $E$, and connect $M E$ and $M N$. Then
$$
\begin{array}{l}
\angle A B C=90^{\circ}-\angle B A D=\angle A E M=\angle A N M \\
\text { Also, since } \angle B A C= \\
\angle N A M, \\
\therefore \triangle A B C \backsim
\end{array}
$$
$\triangle N A M$.
$$
\begin{aligned}
\therefore A C & =\frac{A B \cdot A M}{A N} \\
& =\frac{m c}{n} .
\end{aligned}
$$