2. As shown in Figure 1, $M$ is the midpoint of side $BC$ of $\triangle ABC$, $AN$ bisects $\angle BAC$, $BN \perp AN$ at point $N$, and $AB=10, BC=15, MN=3$. Then the perimeter of $\triangle ABC$ is ( ).
(A) 38
(B) 39
(C) 40
(D) 41
2.D.

As shown in Figure 4, extend $B N$ to intersect $A C$ at $D$.

Since $A N$ bisects $\angle B A C$, and $B N \perp A N$, we have $\triangle A N B \cong \triangle A N D$. Therefore, $A D=A B=10$, and $N$ is the midpoint of $B D$. Also, since $M$ is the midpoint of $B C$, we have $C D=2 M N=6$. Thus, the perimeter of $\triangle A B C$ is
$$
A B+A C+B C=10+(10+6)+15=41 \text {. }
$$