There are four points $A$, $B$, $C$, $D$ on a straight line, $AB: BC: CD=2: 1: 3$. Circles $\odot O_{1}$ and $\odot O_{2}$ are constructed with $AC$ and $BD$ as diameters, respectively. The two circles intersect at $E$ and $F$. Find $ED: EA$.
(1996, Huanggang Region, Hubei Province, Junior High School Mathematics Competition)
Connect $E B$ 、
$$
\begin{array}{l}
E C, \text { let } \angle E C B \\
=\angle 1, \angle E B C \\
=\angle 2, B C=
\end{array}
$$
a, connect $E O_{2}$. Then
$$
\begin{array}{c}
A C=3 a ; \\
A E=3 a \cdot \sin \angle 1, E C=3 a \cdot \cos \angle 1 .
\end{array}
$$

Given $B D=4 a$, we have
$$
\begin{array}{l}
E D=4 a \cdot \sin \angle 2, B E=4 a \cdot \cos \angle 2 . \\
\because B C=a=\frac{1}{2} \cdot 2 a=C O_{2},
\end{array}
$$
$\therefore E C$ is the median of $\triangle B E O_{2}$.
By the median length theorem, we have
$$
E C^{2}=\frac{1}{2} E B^{2}+\frac{1}{2} E O_{2}^{2}-\frac{1}{4} B O_{2}^{2},
$$

which means $9 a^{2} \cos ^{2} \angle 1=8 a^{2} \cos ^{2} \angle 2+2 a^{2}-a^{2}$,
or $9\left(1-\sin ^{2} \angle 1\right)=8\left(1-\sin ^{2} \angle 2\right)+1$.
Thus, $9 \sin ^{2} \angle 1=8 \sin ^{2} \angle 2$.
Therefore, $\frac{\sin \angle 1}{\sin \angle 2}=\frac{2 \sqrt{2}}{3}$.
Hence, $E D: E A=\frac{4 a \cdot \sin \angle 2}{3 a \cdot \sin \angle 1}=\sqrt{2}$. (The author of this article is a second-year junior high school student)