4. In Rt $\triangle A B C$, $\angle B A C=90^{\circ}$, the perpendicular bisector of side $B C$ intersects $A C$ at point $K$, and the perpendicular bisector of segment $B K$ intersects $A B$ at point $L$. If $C L$ bisects $\angle A C B$, find all possible values of $\angle A B C$ and $\angle A C B$.
4. Consider three cases.
(1) $A C>A B$.

As shown in Figure 7, let
$$
\begin{array}{l}
\angle L B K \\
=\angle L K B=\alpha .
\end{array}
$$

Then $\angle K L A=2 \alpha$,
$$
\begin{array}{l}
\angle L K A \\
=90^{\circ}-\angle K L A=90^{\circ}-2 \alpha .
\end{array}
$$

By $B K=K C$
$$
\begin{aligned}
\Rightarrow & \angle K B C=\angle K C B=\frac{1}{2} \angle B K A \\
& =\frac{1}{2}\left(\left(90^{\circ}-\alpha\right)+2 \alpha\right)=45^{\circ}+\frac{1}{2} \alpha .
\end{aligned}
$$

Draw $L T \perp B C$, with the foot of the perpendicular at $T$.
Since $C L$ bisects $\angle A C B$, we have $L T=L A$.
Also, $L B=L K, \angle B T L=\angle K A L=90^{\circ}$
$\Rightarrow \triangle B T L \cong \triangle K A L$
$\Rightarrow \angle L B T=\angle L K A$
$\Rightarrow \alpha+\left(45^{\circ}-\frac{\alpha}{2}\right)=90^{\circ}-2 \alpha$
$\Rightarrow \alpha=18^{\circ}$
$\Rightarrow \angle A B C=45^{\circ}+\frac{1}{2} \alpha=54^{\circ}$
$\Rightarrow \angle A C B=90^{\circ}-\angle A B C=36^{\circ}$.
(2) $A C<A B$.

As shown in Figure 8, let
$\angle L B K$
$=\angle L K B=\alpha$.
Then $\angle K L A=2 \alpha$,
$\angle L K A$
$=90^{\circ}-\angle K L A$
$=90^{\circ}-2 \alpha$.
Draw $L T \perp B C$, with the foot of the perpendicular at $T$.
Since $C L$ bisects $\angle A C B$, we have $L T=L A$.
By $L B=L K, \angle B T L=\angle K A L=90^{\circ}$
$$
\begin{array}{l}
\Rightarrow \triangle B T L \cong \triangle K A L \\
\Rightarrow \angle L B T=\angle L K A=90^{\circ}-2 \alpha \\
\Rightarrow \angle C K B=\angle C B K=90^{\circ}-\alpha \\
\Rightarrow C B=C K .
\end{array}
$$

Also, $B K=C K$, so $\triangle B K C$ is an equilateral triangle.
Hence $\angle C B K=90^{\circ}-\alpha=60^{\circ} \Rightarrow \alpha=30^{\circ}$
$$
\begin{aligned}
\Rightarrow \angle A B C & =90^{\circ}-2 \alpha=30^{\circ}, \\
\angle A C B & =\angle K C B=60^{\circ} .
\end{aligned}
$$
(3) $A C=A B$.

In this case, point $K$ coincides with $A$, and $L$ is the midpoint of side $A B$. Thus, $C L$ is both the median and the angle bisector of $\triangle A B C$, which is impossible.