3. In $\triangle A B C$, $B D$ bisects $\angle A B C$ and intersects $A C$ at point $D, C E$ bisects $\angle A C B$ and intersects $A B$ at point $E$. If $B E+$ $C D=B C$, then the degree measure of $\angle A$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$
3.C.

As shown in Figure 5, let $B D$ and $C E$ intersect at point $O$. Take a point $F$ on $B C$ such that $B F = B E$, and connect $O F$.

From the given information, we know $\angle 1 = \angle 2$, and $B O = B O$, therefore,
$$
\begin{array}{l}
\triangle B E O \cong \triangle B F O, \\
\angle E O B = \angle F O B.
\end{array}
$$

Also, from the given information, $C F = C D$, $\angle 3 = \angle 4$, and $O C = O C$, thus
$\triangle O F C \cong \triangle O D C$, $\angle C O F = \angle C O D$.
Since $\angle B O E = \angle C O D$, it follows that $\angle B O E = \angle B O F = \angle C O F = \angle C O D = \theta$. Therefore, $180^{\circ} = \angle E O C = 3 \theta$.
Thus, $\theta = 60^{\circ}$, and $\angle B O C = 120^{\circ}$.
Also, $\angle B O C = 90^{\circ} + \frac{1}{2} \angle A$, so $\angle A = 60^{\circ}$.