In $\triangle A B C$, $A B=A C, \angle C A B$ and $\angle A B C$'s internal angle bisectors intersect the sides $B C$ and $C A$ at points $D$ and $E$ respectively. Let $K$ be the incenter of $\triangle A D C$. If $\angle B E K=45^{\circ}$, find all possible values of $\angle C A B$. ${ }^{[1]}$
As shown in Figure 1, let $A D$ and $B E$ intersect at point $I$, and the inradius of the incircle $\odot I$ of $\triangle A B C$ be $r$,
$$
\begin{array}{l}
\angle A B C \\
=2 \alpha\left(0<\alpha<45^{\circ}\right) .
\end{array}
$$

Since $A B=A C$, we have
$$
A D \perp B C, \angle A C B=2 \alpha \text {. }
$$
Therefore, $I D=r$,
$$
I E=\frac{r}{\sin \angle B E C}=\frac{r}{\sin 3 \alpha} .
$$

Connecting $D K$. Since $K$ is the incenter of $\triangle A C D$, we know that $I, K, C$ are collinear, and
$$
\begin{array}{l}
\angle I C D=\angle I C E=\alpha, \\
\angle C D K=\angle I D K=45^{\circ}=\angle B E K .
\end{array}
$$

Thus, the circumradii of $\triangle I D K$ and $\triangle I E K$ are equal, i.e.,
$$
\begin{array}{l}
\frac{I D}{\sin \angle I K D}=\frac{I E}{\sin \angle I K E} . \\
\text { Also, } \angle I K D=\angle I C D+\angle C D K=\alpha+45^{\circ}, \\
\angle I K E=\angle B I C-\angle B E K \\
=2\left(90^{\circ}-\alpha\right)-45^{\circ}=135^{\circ}-2 \alpha .
\end{array}
$$

Hence, from equations (1) and (2), we get
$$
\sin \left(\alpha+45^{\circ}\right)=\sin 3 \alpha \cdot \sin \left(135^{\circ}-2 \alpha\right),
$$

which simplifies to
$$
\begin{array}{l}
\sin \alpha+\cos \alpha=\sin 3 \alpha \cdot(\sin 2 \alpha+\cos 2 \alpha) \\
\Leftrightarrow 2 \sin \alpha+2 \cos \alpha \\
\quad=\cos \alpha-\cos 5 \alpha+\sin \alpha+\sin 5 \alpha \\
\Leftrightarrow \sin \alpha+\cos \alpha=\sin 5 \alpha-\cos 5 \alpha \\
\Leftrightarrow \sin \left(\alpha+45^{\circ}\right)=\sin \left(5 \alpha-45^{\circ}\right) \\
\Leftrightarrow 2 \sin \left(2 \alpha-45^{\circ}\right) \cdot \cos 3 \alpha=0 \\
\Leftrightarrow 3 \alpha=90^{\circ} \text { or } 2 \alpha=45^{\circ} \\
\Leftrightarrow \angle B A C=60^{\circ} \text { or } 90^{\circ} .
\end{array}
$$