2. As shown in Figure $1, D$ and $E$ are on the extensions of sides $B A$ and $B C$ of equilateral $\triangle A B C$, respectively, with $A D=A C+$ $C E, \angle B D C=10^{\circ}$. Then the degree measure of $\angle C D E$ is ( ).
(A) $30^{\circ}$
(B) $40^{\circ}$
(C) $45^{\circ}$
(D) $35^{\circ}$
2. B.

As shown in Figure 4, extend $B E$ to $F$ such that $E F = B C$. Connect $D F$. According to the problem, $\triangle B D F$ is an equilateral triangle. Therefore,
$$
\begin{array}{l}
\angle B D F = \angle F = 60^{\circ}, \\
B D = D F, \\
\angle B = \angle F .
\end{array}
$$

Thus, $\triangle B C D \cong \triangle F E D$.
Hence, $\angle E D F = \angle B D C = 10^{\circ}$.
Then, $\angle C D E = \angle B D F - \angle B D C - \angle E D F$
$$
= 60^{\circ} - 10^{\circ} - 10^{\circ} = 40^{\circ} \text{. }
$$