Example 4 As shown in Figure 10, in isosceles right $\triangle ABC$, $\angle A=90^{\circ}, AB$ $=1, D$ is the midpoint of $BC$, and $E, F$ are two other points on side $BC$. $M$ is the other intersection point of the circumcircle of $\triangle ADE$ and the circumcircle of $\triangle ABF$, $N$ is the other intersection point of line $AF$ and the circumcircle of $\triangle ACE$, and $P$ is the other intersection point of line $AD$ and the circumcircle of $\triangle AMN$. Find the length of $AP$.
(2003, China National Training Team Test)
Proof: With $A$ as the inversion center and $r=1$ as the inversion radius, perform an inversion transformation. Under the inversion transformation, denote the image of point $X$ as $X^{\prime}$.
By the properties of inversion,
$B, F, D, E, C$ are collinear
$\Leftrightarrow A, B^{\prime}, F^{\prime}, D^{\prime}, E^{\prime}, C^{\prime}$ are concyclic;
$M$ is the other intersection point of the circumcircle of $\triangle ADE$ and the circumcircle of $\triangle ABF$
$\Leftrightarrow M^{\prime}$ is the intersection of $D^{\prime}E^{\prime}$ and $B^{\prime}F^{\prime}$;
$N$ is the other intersection point of line $AF$ and the circumcircle of $\triangle ACE$
$\Leftrightarrow N^{\prime}$ is the intersection of $AF^{\prime}$ and $C^{\prime}E^{\prime}$;
$P$ is the other intersection point of line $AD$ and the circumcircle of $\triangle AMN$
$\Leftrightarrow P^{\prime}$ is the intersection of $AD^{\prime}$ and $M^{\prime}N^{\prime}$.
As shown in Figure 11, let $B^{\prime}C^{\prime}$ intersect $AD^{\prime}$ at $O^{\prime}$. For the cyclic hexagon $AF^{\prime}B^{\prime}C^{\prime}E^{\prime}D^{\prime}$, by Pascal's theorem, the intersection of $AF^{\prime}$ and $C^{\prime}E^{\prime}$, and the intersection of $D^{\prime}E^{\prime}$ and $B^{\prime}F^{\prime}$ are collinear, i.e., $M^{\prime}, N^{\prime}, O^{\prime}$ are collinear.

Since $P^{\prime}$ is the intersection of $M^{\prime}N^{\prime}$ and $AD^{\prime}$, then $P^{\prime}=O^{\prime}$, i.e., $P^{\prime}$ lies on line $B^{\prime}C^{\prime}$. By the properties of inversion, $A, B, P, C$ are concyclic. Therefore,
$$
\begin{array}{l}
AD \cdot DP = BD \cdot DC \\
\Rightarrow AD = BD = DC = DP = \frac{\sqrt{2}}{2}.
\end{array}
$$

Thus, $AP = \sqrt{2}$.