3. In $\triangle A B C$, $A D$ is the median on side $B C$, points $M$ and $N$ are on sides $A B$ and $A C$ respectively, and satisfy $\angle M D N = 90^{\circ}$. If $B M^{2} + C N^{2} = D M^{2} + D N^{2}$, then the relationship between $A D^{2}$ and $A B^{2} + A C^{2}$ is ( ).
(A) $A D^{2} > A B^{2} + A C^{2}$
(B) $A D^{2} < A B^{2} + A C^{2}$
(C) $A D^{2} = A B^{2} + A C^{2}$
(D) The size of $A D^{2}$ and $A B^{2} + A C^{2}$ is uncertain
3. B.

As shown in Figure 5, draw a line through point $B$ parallel to $A C$ and let it intersect the extension of $N D$ at point $E$. Connect $M E$. Since $B D = D C$, we know $E D = D N$, thus $\triangle B E D \cong \triangle C N D$. Therefore, $B E = C N$. Clearly, $M D$ is the perpendicular bisector of $E N$, so
$$
E M = M N \text {. }
$$
$$
\begin{array}{l}
\text { By } B M^{2} + B E^{2} = B M^{2} + C N^{2} \\
= D M^{2} + D N^{2} = M N^{2} = E M^{2},
\end{array}
$$

we know $\triangle B E M$ is a right triangle, $\angle M B E = 90^{\circ}$. 
Therefore, $\angle A B C + \angle A C B = \angle A B C + \angle E B C = 90^{\circ}$. 
Thus, $\angle B A C = 90^{\circ}$. 
Hence, $A D^{2} = \left(\frac{1}{2} B C\right)^{2} = \frac{1}{4}\left(A B^{2} + A C^{2}\right)$.