4. Take line segment $AB$ as the diameter to draw a semicircle, with the center at $O$. $C$ is a point on the semicircle, and $OC^2 = AC \cdot BC$. Then $\angle CAB=$ $\qquad$
4. $15^{\circ}$ or $75^{\circ}$. As

shown in the figure, construct $C D \perp A B$, with
D being the foot of the perpendicular. By the area formula and the given

information, we have $C D=\frac{1}{2} O C$,
$$
\angle C O D=30^{\circ} \text {. }
$$

If point $D$ is on line segment $O B$, since $\triangle A O C$ is an isosceles triangle, $\angle C O D$ is an exterior angle of $\triangle A O C$, hence $\angle C A B=15^{\circ}$; if point $D$ is on line segment $O A$, then $\angle C B A=15^{\circ}, \angle C A B=75^{\circ}$.