4. In an acute triangle $ABC$, $\angle A=30^{\circ}$. A circle is drawn with $BC$ as its diameter, intersecting $AB$ and $AC$ at points $D$ and $E$ respectively. Connecting $D$ and $E$, the triangle $ABC$ is divided into triangle $ADE$ and quadrilateral $BDEC$. Let the areas of these shapes be $S_{1}$ and $S_{2}$ respectively, then $S_{1}: S_{2}=$ $\qquad$ .
4. 3

As shown in the figure, $BC$ is the diameter of the circle,
$$
\begin{array}{l}
\angle AEB = 180^{\circ} - \angle BEC = 90^{\circ}, \\
\therefore \quad \frac{AE}{AB} = \cos A = \cos 30^{\circ} = \frac{\sqrt{3}}{2}. \\
\text{Also, } \triangle ADE \sim \triangle ABC, \\
\therefore \quad \frac{AD}{AC} = \frac{AE}{AB} = \frac{\sqrt{3}}{2}.
\end{array}
$$

Therefore,
$$
\begin{aligned}
\frac{S_{\triangle AED}}{S_{\triangle ABC}} & = \frac{\frac{1}{2} AD \cdot AE \sin A}{\frac{1}{2} AB \cdot AC \sin A} \\
& = \left(\frac{AE}{AB}\right)^2 = \frac{3}{4}.
\end{aligned}
$$

Thus, the area of quadrilateral $DBCE$
$$
S_2 = \frac{1}{4} S_{\triangle ABC}, \quad \therefore \quad S_1 : S_2 = 3
$$