4. Given a line segment $A B$ as the diameter of a semicircle, with the center at $O$, and $C$ is a point on the semicircle. If $O C^{2}=A C \cdot B C$, then $\angle C A B=$ $\qquad$ .
4. $75^{\circ}$ pear $155^{\circ}$

Draw $\% A B$, with the foot of the perpendicular at D, we have
$A C \cdot B C=O C^{2}=$ $\frac{1}{4} A B^{2}$,
On the other hand, by the area formula
$$
A C \cdot B C=C D \cdot A B \text {, }
$$

Therefore, $C D \cdot A B=\frac{1}{4} A B^{2}$, which means $C D=\frac{1}{4} A B$, thus $C D=\frac{1}{2} O C$. Suppose $D$ is on the segment $A O$. Then $\angle C O D=$ $30^{\circ}$. Since $\triangle C O B$ is an isosceles triangle, $\angle C O D$ is one of its exterior angles, hence $\angle C B A=15^{\circ}$, thus $\angle C A B=75^{\circ}$.

If $D$ is on the segment $O B$, then by the same discussion, $\angle C A B=15^{\circ}$.