1. Given that $A D$ and $B E$ are the angle bisectors of $\triangle A B C$, and they intersect at point $I$, with $I D=I E$. If $\angle C A B=80^{\circ}$, then $\angle A B C=$ $\qquad$
II. 1. $40^{\circ}$ or $80^{\circ}$.
It is easy to know that $I$ is the incenter, $\angle I C E = \angle I C D$.
From the given, $I E = I D$.
Consider $\triangle C I E$ and $\triangle C I D$:
If $C E = C D$, then $\angle C E I = \angle C D I$;
If $C E \neq C D$, then $\angle C E I + \angle C D I = 180^{\circ}$.
Therefore, $\angle C A B = \angle C B A$ or $\angle C A B + \angle C B A = 120^{\circ}$.
Given that $\angle C A B = 80^{\circ}$, thus,
$\angle A B C = 40^{\circ}$ or $80^{\circ}$.