3. As shown in Figure 2, fold and press the paper piece $\triangle A B C$ along $D E$. Then $\angle A=(\quad)$.
(A) $\angle 1+\angle 2$
(B) $\frac{1}{2}(\angle 1+\angle 2)$
(C) $\frac{1}{3}(\angle 1+\angle 2)$
(D) $\frac{1}{4}(\angle 1+\angle 2)$
3. B.
From the sum of the interior angles of a quadrilateral being $360^{\circ}$, we get
$$
\begin{array}{l}
\angle B+(\angle 1+\angle A D E)+(\angle A E D+\angle 2)+\angle C \\
=360^{\circ} .
\end{array}
$$

And $\angle B+\angle C=\angle A D E+\angle A E D=180^{\circ}-\angle A$, thus
$$
\angle A=\frac{1}{2}(\angle 1+\angle 2) \text {. }
$$