4. As shown in Figure 1, in the right triangle $\triangle A B C$, $\angle C=60^{\circ}$. With $C$ as the center and $B C$ as the radius, a circle intersects $A C$ at point $D$. Connect $B D$. The arc $\overparen{B D}$ and the chord $B D$ divide $\triangle A B C$ into three parts. Then the ratio of the areas of the three parts $S_{1}: S_{2}$ :
$$
S_{3}=
$$
4. $\left(\sqrt{3}-\frac{\pi}{3}\right):\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right): \frac{\sqrt{3}}{2}$.

Let $BC = x$. Then
$$
\begin{array}{l}
S_{1}=\frac{1}{2} AB \cdot BC - \frac{1}{2} BC \cdot \overparen{BD} \\
=\frac{1}{2} x^{2}\left(\tan \frac{\pi}{3} - \frac{\pi}{3}\right) = \frac{1}{2}\left(\sqrt{3} - \frac{\pi}{3}\right) x^{2}, \\
S_{2}=\frac{1}{2} BC \cdot \overparen{BD} - \frac{1}{2} BC \cdot CD \sin \frac{\pi}{3} \\
=\frac{1}{2}\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) x^{2}, \\
S_{3}=\frac{1}{2} BC \cdot CD \sin \frac{\pi}{3} = \frac{1}{2} \times \frac{\sqrt{3}}{2} x^{2} .
\end{array}
$$

Therefore, $S_{1}: S_{2}: S_{3} = \left(\sqrt{3} - \frac{\pi}{3}\right):\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right): \frac{\sqrt{3}}{2}$.