Example 3 As shown in Figure 3, given that quadrilateral $ABCD$ is a square, $\odot O$ passes through vertex $A$ of the square and the intersection point $P$ of the diagonals, and intersects $AB, AD$ at points $F, E$ respectively.
(1) Prove:
$$
DE = AF \text{; }
$$
(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)
Solution: Clearly, $EF$ is the diameter of $\odot O$, meaning point $O$ lies on side $EF$.

Connecting $EP$ and $FP$, it is easy to see that $\triangle EPF$ is an isosceles right triangle. Therefore, $EP = \sqrt{2} OE = \frac{\sqrt{6}}{2}$.
(1) Since $DP = AP$, $\angle EDP = 45^{\circ} = \angle FAP$, and $\angle DEP = \angle AFP$, we have
$\triangle DEP \cong \triangle AFP \Rightarrow DE = AF$.
(2) Draw $PM \perp AD$ at point $M$, then $M$ is the midpoint of side $AD$.
Thus, $AM = \frac{\sqrt{2} + 1}{2}$, $AP = \sqrt{2} AM = \frac{\sqrt{2} + 2}{2}$.
Let $AE = x$. Then $EM = \frac{\sqrt{2} + 1}{2} - x$.
Applying the property to the right triangle $\triangle APM$ and point $E$, we have
$$
AP^2 = AE^2 + EP^2 + 2 EA \cdot EM,
$$

which is $\left(\frac{\sqrt{2} + 2}{2}\right)^2 = x^2 + \left(\frac{\sqrt{6}}{2}\right)^2 + 2x\left(\frac{\sqrt{2} + 1}{2} - x\right)$,
or $x^2 - (\sqrt{2} + 1)x + \sqrt{2} = 0$.
Solving for $AE = x = 1$ or $\sqrt{2}$.
Therefore, $\frac{AE}{ED} = \sqrt{2}$ or $\frac{\sqrt{2}}{2}$.