Example 2. As shown in Figure 2, the area of trapezoid $ABCD$ is $S$,
$$
\begin{array}{l}
AB \| CD, AB=b, \\
CD=a(a<b),
\end{array}
$$

Diagonals $AC$ and $BD$ intersect
at $O$, and the area of $\triangle BOC$ is $\frac{2 S}{9}$. Find $-\frac{a}{b}$.
(1988, Jiangsu Junior
Competition)
Given $\because A B \| C D, \therefore \triangle A O B \backsim \triangle C O D$,
$$
\therefore \frac{O D}{O C}=\frac{O B}{O A} \text {. }
$$

Also, $\frac{S \triangle \frac{C O D}{}}{2 S / 9}=\frac{O D}{O B}, \frac{2 S / 9}{S \triangle A O B}=\frac{O C}{O A}$,
$\therefore \frac{S \triangle C O D}{2 S / 9}=\frac{2 S / 9}{S \triangle \triangle O B}$,
$\therefore S_{\triangle A O B} \cdot S_{\triangle C O D}=\left(\frac{2 S}{9}\right)^{2}$.
Also, $S_{\triangle A O D}=S_{\triangle B O C}=-\frac{2 S}{9}$,
$\therefore S \therefore \triangle O B+S \triangle C O D=\frac{5}{9} S$.
From (1) and (2), we know that $S_{\triangle A O B}, S_{\triangle C O D}$ are the roots of the equation $t^{2}-\frac{5}{9} s t+\left(\frac{2 s}{9}\right)^{2}=0$, noting that $S_{\triangle A O B}>S_{\triangle C O D}$, solving gives $S_{\triangle A O B}=\frac{4}{9} S$, $S_{\triangle C \cap D}=\frac{S}{9}$,
thus $\left(\frac{a}{b}\right)^{2}=\frac{S}{9} / \frac{4}{9} S=1 / 4$.

Therefore, $a: b=1: 2$.