2. Let $A D$ be the angle bisector of $\angle B A C$, $A B=8, A C$ $=10, A D=6, E$ be a point on $A C$, $A E=2, M$ be the midpoint of $A E$, $N$ be the midpoint of $B C$. Then $M N=$ ( ).
(A) 5
(B) $\frac{26}{5}$
(C) $\frac{27}{5}$
(D) $\frac{11}{2}$
2. C.

As shown in Figure 4, connect $B E$, take the midpoint $P$ of $B E$, and connect $P M \backslash P N$.
$$
\begin{array}{l}
\text { Then } P M \Perp \frac{1}{2} A B, \\
P N \Perp \frac{1}{2} C E .
\end{array}
$$

Therefore, $A B=C E=8 \Rightarrow P M=P N$
$$
\begin{array}{l}
\Rightarrow \angle P M N=\angle P N M=\angle C M N . \\
\text { Also, } \angle C A D=\frac{1}{2} \angle B A C=\frac{1}{2} \angle P M C=\angle C M N \\
\Rightarrow M N \parallel A D \Rightarrow \frac{M N}{A D}=\frac{M C}{A C} \\
\Rightarrow M N=\frac{A D \cdot M C}{A C}=\frac{27}{5} .
\end{array}
$$