Example 2 In the acute triangle $\triangle ABC$, $\angle A=30^{\circ}$, a circle is constructed with $BC$ as its diameter, intersecting $AB$ and $AC$ at $D$ and $E$ respectively. Connecting $DE$ divides $\triangle ABC$ into $\triangle ADE$ and quadrilateral $DBCE$, with areas denoted as $S_{1}$ and $S_{2}$ respectively. Then $S_{1}: S_{2}=$ $\qquad$
: As shown in Figure 2, connect $B E$, then $\angle A E B=90^{\circ}$.
$$
\begin{array}{l}
\therefore \frac{A E}{A B}=\cos 30^{\circ}=\frac{\sqrt{3}}{2} . \\
\because \angle A E D=\angle A B C, \\
\therefore \triangle A E D \operatorname{\triangle o} \triangle A B C . \\
\therefore \frac{S_{\triangle A E D}}{S_{\triangle A B C}}=\left(\frac{A E}{A B}\right)^{2} \\
=\frac{3}{4},
\end{array}
$$

That is, $S_{1}=\frac{3}{4} S_{\triangle A B C}$.
Thus, $S_{2}=\frac{1}{4} S_{\triangle A B C}$.
$$
\therefore S_{1}: S_{2}=3: 1 \text {. }
$$