2. In quadrilateral $ABCD$, $AC$ bisects $\angle BCD$, $\angle CAD=20^{\circ}, \angle CBD=40^{\circ}$. Then $\angle ABD=(\quad)$.
(A) $50^{\circ}$
(B) $60^{\circ}$
(C) $70^{\circ}$
(D) $80^{\circ}$
$$
\begin{array}{l} 
\angle C A E = \angle C B E \\
= 40^{\circ}, \\
\angle D A E \\
= \angle C A E - \angle C A D \\
= 20^{\circ} = \angle C A D, \\
\angle A E B = \angle A C B = \angle A C D . \\
\text { Also, } A D = A D, \text { so } \\
\triangle A D E \cong \triangle A D C \\
\Rightarrow A E = A C \Rightarrow \overparen{A E} = \overparen{A C} \\
\Rightarrow \angle A B E + \angle A B C = 180^{\circ} \\
\Rightarrow \angle A B D = \angle A B E = 70^{\circ} .
\end{array}
$$