16. As shown in Figure 9, quadrilateral $ABCD$ is a square, $\odot O$ passes through the vertex $A$ and the intersection point $P$ of the diagonals, and intersects $AB$ and $AD$ at points $F$ and $E$, respectively.
(1) Prove that $DE = AF$;
(2) If the radius of $\odot O$ is $\frac{\sqrt{3}}{2}$ and $AB = \sqrt{2} + 1$, find the value of $\frac{AE}{ED}$.
16. (1) As shown in Figure 12, connect \( P E, P F, E F \).

Since \(\angle E A F = 90^{\circ}\), \( E F \) is the diameter of \(\odot O\).
Thus, \(\angle F P E = 90^{\circ}\).
Also, \(\angle A P D = 90^{\circ}\), so \(\angle E P D = \angle A P F\).
Clearly, \( P D = P A \),
\(\angle P A F = \angle P D E = 45^{\circ}\).
Therefore, \(\triangle P D E \cong \triangle P A F\).
Hence, \( D E = A F \).
(2) Since \( D E = A F \), we have
\[
A E + A F = A D = \sqrt{2} + 1.
\]

Also, \( A E^{2} + A F^{2} = E F^{2} = 3 \), which means
\[
(A E + A F)^{2} - 2 A E \cdot A F = 3.
\]

Thus, \( A E \cdot A F = \sqrt{2} \).
Therefore, \( A E \) and \( A F \) are the roots of the equation \( x^{2} - (\sqrt{2} + 1) x + \sqrt{2} = 0 \).
Solving this, we get \( A E = \sqrt{2}, A F = 1 \) or \( A E = 1, A F = \sqrt{2} \).
Hence, \(\frac{A E}{E D} = \sqrt{2}\) or \(\frac{\sqrt{2}}{2}\).