Example 2.3.7 (2009, IMO50) In $\triangle ABC$, $AB=AC$, the internal angle bisectors of $\angle CAB$ and $\angle ABC$ intersect the sides $BC$ and $CA$ at points $D$ and $E$, respectively. Let $K$ be the incenter of $\triangle ADC$. If $\angle BEK=45^{\circ}$, find all possible values of $\angle CAB$.
Prove that line segments $A D$ and $B E$ are angle bisectors of $\triangle A B C$, and they intersect at the incenter of $\triangle A B C$. Connect $C I$. Then $C I$ bisects $\angle A C B$.
Since $K$ is the incenter of $\triangle A D C$, point $K$ lies on line segment $C I$.
Let $\angle B A C=\alpha$. Since $A B=A C$, we have $A D \perp B C$. Therefore, $\angle A B C=\angle A C B=90^{\circ}-\frac{\alpha}{2}$.
Since $B I$ and $C I$ bisect $\angle A B C$ and $\angle A C B$ respectively, we have
$$
\angle A B I=\angle I B C=\angle A C I=\angle I C B=45^{\circ}-\frac{\alpha}{4} \text {. }
$$

Thus, $\angle E I C=\angle I B C+\angle I C B=90^{\circ}-\frac{\alpha}{2}$.
$\angle I E C=\angle B A E+\angle A B E=45^{\circ}+\frac{3 \alpha}{4} . \angle B E K=45^{\circ}, \angle K E C=\frac{3 \alpha}{4}$. Therefore,
$$
\begin{aligned}
I K: K C= & S_{\triangle I E K}: S_{\triangle E K C}=\frac{1}{2} I E \cdot E K \sin \angle I E K: \frac{1}{2} E C \cdot E K \sin \angle K E C \\
& =\left(\sin 45^{\circ}: \sin \frac{3 \alpha}{4}\right)(I E: E C) \\
& =\left(\sin 45^{\circ}: \sin \frac{3 \alpha}{4}\right)\left(\sin \left(45^{\circ}-\frac{\alpha}{4}\right): \sin \left(90^{\circ}-\frac{\alpha}{2}\right)\right)
\end{aligned}
$$

On the other hand, since $K$ is the incenter of $\triangle A D C$, $D K$ bisects $\angle I D C$. By the angle bisector theorem,
$$
I K: K C=I D: D C=\tan \angle I C D=\sin \left(45^{\circ}-\frac{\alpha}{4}\right): \cos \left(45^{\circ}-\frac{\alpha}{4}\right) .
$$

Clearing the denominators, we get $2 \sin 45^{\circ} \cdot \cos \left(45^{\circ}-\frac{\alpha}{4}\right)=2 \sin \frac{3 \alpha}{4} \cos \frac{\alpha}{2}$.
Using the product-to-sum formulas, we get $\sin \left(90^{\circ}-\frac{\alpha}{4}\right)+\sin \frac{\alpha}{4}=\sin \frac{5 \alpha}{4}+\sin \frac{\alpha}{4}$, which simplifies to $\sin \left(90^{\circ}-\frac{\alpha}{4}\right)=\sin \frac{5 \alpha}{4}$.
Since $00$. Therefore, $\sin \frac{5 \alpha}{4}>0$. Thus, $0<\frac{5 \alpha}{4}<180^{\circ}$. Hence, only $90^{\circ}-\frac{\alpha}{4}=\frac{5 \alpha}{4} \longrightarrow \alpha=60^{\circ}$.
Or $90^{\circ}-\frac{\alpha}{4}=180^{\circ}-\frac{5 \alpha}{4} \longrightarrow \alpha=90^{\circ}$.
When $\alpha=60^{\circ}$, it is easy to verify that $\triangle I E C \cong \triangle I D C$. Therefore, $\triangle I E K \cong \triangle I D K$.
Thus, $\angle B E K=\angle I D K=45^{\circ}$.
When $\alpha=90^{\circ}$, $\angle E I C=90^{\circ}-\frac{\alpha}{2}=45^{\circ}=\angle K D C$.
Also, $\angle I C E=\angle D C K$, so $\triangle I C E \backsim \triangle D C K$. This implies $I C \cdot K C=D C \cdot E C$. Therefore, $\triangle I D C \sim \triangle E K C$. Thus, $\angle E K C=\angle I D C=90^{\circ}$.
Hence, $\angle B E K=180^{\circ}-\angle E I K-\angle E K I=45^{\circ}$.
In summary, the possible values of $\angle C A B$ are $60^{\circ}$ and $90^{\circ}$.