4. In $\triangle A B C$, $A B=A C$, the internal angle bisectors of $\angle C A B$ and $\angle A B C$ 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 $\angle C A B$. (Belgium provided)
4. Segments $A D$ and $B E$ are the angle bisectors of $\triangle A B C$, intersecting at the incenter $I$ of $\triangle A B C$. Connecting $C I$. Then $C I$ bisects $\angle A C B$.

Since $K$ is the incenter of $\triangle A D C$, point $K$ lies on segment $C I$.
Let $\angle B A C=\alpha$.
Since $A B=A C$, we have $A D \perp B C$, so
$\angle A B C=\angle A C B=90^{\circ}-\frac{\alpha}{2}$.
Since $B I$ and $C I$ bisect $\angle A B C$ and $\angle A C B$ respectively, we have $\angle A B I=\angle I B C=\angle A C I$
$$
=\angle I C B=45^{\circ}-\frac{\alpha}{4} \text {. }
$$

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

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,
$\frac{I K}{K C}=\frac{I D}{D C}=\tan \angle I C D=\frac{\sin \left(45^{\circ}-\frac{\alpha}{4}\right)}{\cos \left(45^{\circ}-\frac{\alpha}{4}\right)}$.
Thus, $\frac{\sin 45^{\circ}}{\sin \frac{3 \alpha}{4}} \cdot \frac{\sin \left(45^{\circ}-\frac{\alpha}{4}\right)}{\sin \left(90^{\circ}-\frac{\alpha}{2}\right)}=\frac{\sin \left(45^{\circ}-\frac{\alpha}{4}\right)}{\cos \left(45^{\circ}-\frac{\alpha}{4}\right)}$.
Clearing the denominators, we get
$2 \sin 45^{\circ} \cdot \cos \left(45^{\circ}-\frac{\alpha}{4}\right)=2 \sin \frac{3 \alpha}{4} \cdot \cos \frac{\alpha}{2}$.
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, $\sin \frac{5 \alpha}{4}>0$, i.e., $0<\frac{5 \alpha}{4}<180^{\circ}$.
Thus, we have $90^{\circ}-\frac{\alpha}{4}=\frac{5 \alpha}{4} \Rightarrow \alpha=60^{\circ}$,

or $90^{\circ}-\frac{\alpha}{4}=180^{\circ}-\frac{5 \alpha}{4} \Rightarrow \alpha=90^{\circ}$.
When $\alpha=60^{\circ}$,
it is easy to verify that $\triangle I E C \cong \triangle I D C$.
Therefore, $\triangle I E K \cong \triangle I D K$.
Hence, $\angle B E K=\angle I D K=45^{\circ}$.
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 $\triangle I C E \backsim \triangle D C K$, which implies $I C \cdot K C=D C \cdot E C$.
Thus, $\triangle I D C \backsim \triangle E K C$.
Therefore, $\angle E K C=\angle I D C=90^{\circ}$.
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 $90^{\circ}$.