Example 3 The radius of the circle is both the mean proportional between two chords and equal to the difference of these two chords. Find the degree measures of the central angles subtended by the two chords.

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Explanation: As shown in Figure 2, $AB$ and $CD$ are two chords of $\odot O$, with $AB > CD$. Take $AM = CD$, then $BM = OB'$.
From the given information,
$$
\begin{array}{l}
OA^2 = AB \cdot CD \\
= AB \cdot AM,
\end{array}
$$

we get $\frac{OA}{AB} = \frac{AM}{OA}$.
Therefore, $\triangle OAB \sim \triangle MAO$.
Thus, $OM = AM = CD$.
Hence, $\triangle BOM \cong \triangle OCD$.
So, $\angle COD = \angle B$.
Let $\angle A = \angle B = \alpha$, then
$$
\begin{array}{l}
\angle AOM = \angle A = \alpha, \\
\angle BOM = \angle BMO = \angle AOM + \angle A = 2\alpha, \\
\angle AOB = \angle AOM + \angle BOM = 3\alpha.
\end{array}
$$

In $\triangle OAB$, since
$$
\angle AOB + \angle A + \angle B = 3\alpha + \alpha + \alpha = 180^\circ,
$$

we have $\alpha = 36^\circ$.
Therefore, $\angle AOB = 3\alpha = 108^\circ$,
$$
\angle COD = \alpha = 36^\circ.
$$