2. $126^{\circ}$.

Since $B C / / A D, C A$ bisects $\angle B C D$, so,
$$
\begin{array}{l}
\angle D A C=\angle A C B=\angle A C D \\
\Rightarrow D A=D C=A O \\
\Rightarrow \angle A D O=\angle A O D .
\end{array}
$$
$$
\begin{array}{l}
\text { Let } \angle D A C=\angle A C B=\angle A C D=\alpha, \\
\angle A D O=\angle A O D=\beta .
\end{array}
$$

By $B C / / A D$, we know $\triangle A D O \backsim \triangle C B O$.
Combining $A D=A O$, we get
$$
O C=B C \text {, and } \angle C B O=\angle C O B=\beta \text {. }
$$

Also, $B C=O D$, thus,
$$
O C=O D \Rightarrow \angle O D C=\angle O C D=\alpha \text {. }
$$

Combining the diagram, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ \beta = 2 \alpha , } \\
{ \alpha = 2 \beta = 1 8 0 ^ { \circ } }
\end{array} \Rightarrow \left\{\begin{array}{l}
\alpha=36^{\circ}, \\
\beta=72^{\circ} .
\end{array}\right.\right. \\
\Rightarrow \angle D B C=\angle D C B=72^{\circ} \\
\Rightarrow B D=C D=A D \\
\Rightarrow \angle D A B=\angle D B A=54^{\circ} \\
\Rightarrow \angle A B C=\angle A B D+\angle D B C=126^{\circ} . \\
\end{array}
$$
2. In quadrilateral $A B C D$, $B C / / A D, C A$ bisects $\angle B C D, O$ is the intersection of the diagonals, $C D=A O, B C=$ $O D$. Then $\angle A B C=$ $\qquad$ .