2. As shown in Figure 1, in the right triangle $\triangle ABC$, $\angle A=90^{\circ}, AB=$ $AC, \angle ABC$ is bisected by $BD$, and $DE \perp BC$ at $E$. If the perimeter of $\triangle CDE$ is 6, then the length of $BC$ is ( ).
(A) $6-3 \sqrt{2}$
(B) $12-6 \sqrt{2}$
(C) $6 \sqrt{2}$
(D) 6
2.D.

From $\angle C B D=\angle A B D, \angle D E B=\angle A=90^{\circ}, B D=$ $B D$, we know $\triangle B A D \cong \triangle B E D$. Therefore, $B E=A B=A C, D E=$ $A D$. Thus,
$$
\begin{array}{l}
B C=B E+E C=A C+E C \\
=A D+D C+E C=D E+D C+E C=6 .
\end{array}
$$