3. As shown in Figure 1, in $\triangle A B C$, $A B=A C$, points $D, E$ are on sides $A B, A C$ respectively, $D M$ bisects $\angle B D E$, and $E N$ bisects $\angle D E C$. If $\angle D M N=$ $110^{\circ}$, then $\angle D E A=$ ( ).
(A) $40^{\circ}$
(B) $50^{\circ}$
(C) $60^{\circ}$
(D) $70^{\circ}$
3.A.

Since $AB = AC$, therefore, $\angle B = \angle C$.
$$
\begin{array}{l}
\text{Also, } \angle B = \angle DMN - \angle BDM = \angle DMN - \angle MDE, \\
\angle C = \angle MNE - \angle NEC = \angle MNE - \angle NED,
\end{array}
$$

Thus, $\angle DMN - \angle MDE = \angle MNE - \angle NED$,
which means $\angle DMN + \angle NED = \angle MNE + \angle MDE$.
In quadrilateral DMNE, by
$$
\angle DMN + \angle NED = \angle MNE + \angle MDE,
$$

we have $\angle DMN + \angle NED = \angle MNE + \angle MDE = 180^{\circ}$. Therefore, $\angle NED = 70^{\circ}$.
Hence, $\angle DEA = 180^{\circ} - 2 \angle NED = 40^{\circ}$.