Example 3 As shown in Figure 5, quadrilateral $ABCD$ is a square, $\odot O$ passes through vertex $A$ and the intersection point $P$ of the diagonals, and intersects $AB, AD$ at points $F, E$ respectively.
(1) Prove: $DE = AF$;
(2) If the radius of $\odot O$ is $\frac{\sqrt{3}}{2}$, and $AB = \sqrt{2} + 1$, find the value of $\frac{AE}{ED}$.
(21st Junior High School Mathematics Competition in Jiangsu Province)
Analysis: (1) It is easy to think of connecting $E P$ and $F P$ to prove $\triangle D E P \cong \triangle A F P$. This is because $A B C D$ is a square, so $D P=A P, \angle A D P=\angle B A P=45^{\circ}, \angle E P F=$ $180^{\circ}-\angle B A D=90^{\circ}=\angle A P D$, hence $\angle E P D=$ $\angle A P F$ is proved.
(2) Connect $E F$. Since $\angle B A D=90^{\circ}$, $E F$ is the diameter. Therefore, $E F=\sqrt{3}$.
By the Pythagorean theorem, $A E^{2}+A F^{2}=E F^{2}$.
And $A E+A F=A E+E D=\sqrt{2}+1$, so
$$
A E^{2}+(\sqrt{2}+1-A E)^{2}=3 \text {. }
$$

Solving this, we get $A E=\sqrt{2}$ or 1.
Thus, the corresponding $E D=1$ or $\sqrt{2}$.
Therefore, $\frac{A E}{E D}=\sqrt{2}$ or $\frac{\sqrt{2}}{2}$.