Question 1 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 $\angle C A B$ ${ }^{[1,2]}$.
As shown in Figure 1, if $\angle C E K=45^{\circ}$, then because $\angle B E K=45^{\circ}$, we have $\angle B E C=90^{\circ}$.
Also, $\angle A B E=\angle C B E$, so $A B=B C$.
Therefore, $A B=A C=B C$.
Thus, $\angle B A C=60^{\circ}$.
As shown in Figure 2, if $\angle C E K \neq 45^{\circ}$, then $\angle B E C \neq 90^{\circ}$.
Let $A D$ intersect $B E$ at point $I$. Draw $I F \perp A C$ at point $F$. Connect $D K$ and $F K$.

Since $I$ and $K$ are the incenters of $\triangle A B C$ and $\triangle A C D$ respectively, $I$, $K$, and $C$ are collinear.
Also, since $A B=A C$ and $A D$ bisects $\angle B A C$, we have
$A D \perp B C$.
Since $C I$ is the angle bisector of $\angle A C D$, it is easy to see that
$\triangle I D K \cong \triangle I F K$.
Thus, $\angle I F K=\angle I D K$, and $\angle F K I=\angle D K I$.
Since $K$ is the incenter of $\triangle A C D$, we have
$\angle I D K=\frac{1}{2} \angle A D C=45^{\circ}$.
Therefore, $\angle I F K=45^{\circ}=\angle I E K$.
Hence, $I$, $K$, $E$, and $F$ are concyclic.
Thus, $\angle F E I=\angle F K I=\angle D K I$.
Let $\angle I C D=\alpha$. Then
$\angle I B C=\alpha$, and $\angle E C B=2 \alpha$.
Therefore, $\angle F E I=\angle I B C+\angle E C B=3 \alpha$,
$\angle D K I=\angle I C D+\angle K D C=45^{\circ}+\alpha$.
Thus, $3 \alpha=\alpha+45^{\circ} \Rightarrow 2 \alpha=45^{\circ}$
$\Rightarrow \angle A C B=45^{\circ} \Rightarrow \angle B A C=90^{\circ}$.
In summary, $\angle B A C=60^{\circ}$ or $90^{\circ}$.