2. As shown in Figure 1, let $M$ and $N$ be two points on the diagonal $BD$ of square $ABCD$, with $BM=3$, $DN=4$, and $BD=12$. Then $\angle MAB + \angle NCD$ equals ( ).
(A) $40^{\circ}$
(B) $45^{\circ}$
(C) $48^{\circ}$
(D) $60^{\circ}$
2. B.

As shown in Figure 6, construct the congruent triangle $\triangle A B^{\prime} M$ of $\triangle A B M$, and let the angle bisector of $\angle B^{\prime} A D$ intersect $B D$ at point $P$. Thus,
$$
\triangle A D P \cong \triangle A B^{\prime} P \text {. }
$$

Then $B M=B^{\prime} M$,
$P D=P B^{\prime}$,
$\angle P A M=45^{\circ}$,
$\angle M B^{\prime} P=90^{\circ}$.
Therefore, $B M^{2}=M P^{2}-P D^{2}=M N^{2}-N D^{2}$.
Hence, point $P$ coincides with $N$, $\angle M A N=45^{\circ}$.
By the symmetry of the square,
$$
\angle M A B+\angle N C D=\angle M A B+\angle N A D=45^{\circ}
$$