2. In $\triangle A B C$, $A B>A C, I$ is the incenter of $\triangle A B C$. Let $\Gamma$ denote the circle with diameter $A I$. Suppose circle $\Gamma$ intersects the circumcircle of $\triangle A B C$ at points $A$ and $D$, and $D$ lies on the arc $\overparen{A C}$ not containing point $B$. Draw a line through point $A$ parallel to $B C$, intersecting circle $\Gamma$ at $A$ and another point $E$, as shown in Figure 1. If $D I$ bisects $\angle C D E$, and $\angle A B C=33^{\circ}$, find the measure of $\angle B A C$.
(Xiong Bin, problem contributor)
2. Connect $A D$.

From the conditions, we have
$$
\begin{array}{l}
\angle I D C=\angle I D E=\angle I A E \\
=\angle I A B+\angle B A E \\
=\frac{\angle B A C}{2}+\angle A B C .
\end{array}
$$

Notice that, $\angle I D A=90^{\circ}$.
$$
\begin{array}{l}
\text { Then } \angle I D C=\angle A D C-\angle A D I \\
=\left(180^{\circ}-\angle A B C\right)-90^{\circ} \\
=90^{\circ}-\angle A B C .
\end{array}
$$

Comparing equations (1) and (2), we get
$$
\begin{array}{l}
\frac{\angle B A C}{2}+\angle A B C=90^{\circ}-\angle A B C \\
\Rightarrow \angle B A C=2\left(90^{\circ}-2 \angle A B C\right)=48^{\circ} .
\end{array}
$$