2. As shown in Figure 1, $C D / / B E$. Then $\angle 2+\angle 3-\angle 1=$ ( ).
(A) $90^{\circ}$
(B) $120^{\circ}$
(C) $150^{\circ}$
(D) $180^{\circ}$
2.D. Extend $D C$ to intersect $A B$ at $F$. Since $C D / / B E$, we have $\angle B F C=\angle 3$. Then
$$
\begin{array}{l}
\angle 2-\angle 1=\angle A F C \\
=180^{\circ}-\angle B F C=180^{\circ}-\angle 3 .
\end{array}
$$