14 In the isosceles right $\triangle A B C$, $\angle A=90^{\circ}, A B=1, D$ is the midpoint of $B C$, and $E, F$ are two other points on $B C$. $M$ is the other intersection point of the circumcircle of $\triangle A D E$ and the circumcircle of $\triangle A B F$; $N$ is the other intersection point of line $A F$ and the circumcircle of $\triangle A C E$; $P$ is the other intersection point of line $A D$ and the circumcircle of $\triangle A M N$ (Fig. 4). Find the length of $A P$.
Solve by using inversion.

Let point $A$ be the center of inversion, and $r=1$ be the radius of the inversion. Under this inversion, use $X^{*}$ to represent the image of point $X$. By the properties of inversion: $B, F, D, E, C$ are collinear $\Leftrightarrow A, B^{*}, F^{*}, D^{*}, E^{*}, C^{*}$ are on the same circle; $M$ is the other intersection of the circumcircles of $\triangle ADE$ and $\triangle ABF$ $\Leftrightarrow M^{*}$ is the intersection of $D^{*}E^{*}$ and $B^{*}F^{*}$; $N$ is the other intersection of $AF$ and the circumcircle of $\triangle ACE$ $\Leftrightarrow N^{*}$ is the intersection of $AF^{*}$ and $C^{*}E^{*}$; $P$ is the other intersection of $AD$ and the circumcircle of $\triangle AMN$ $\Leftrightarrow P^{*}$ is the intersection of $AD^{*}$ and $M^{*}N^{*}$. (Figure 5)

Let $B^{*}C^{*}$ intersect $AD^{*}$ at $O^{*}$. Applying Pascal's theorem to the cyclic hexagon $AF^{*}B^{*}C^{*}E^{*}D^{*}$, we know:
The intersection of $AF^{*}$ and $C^{*}E^{*}$, the intersection of $F^{*}B^{*}$ and $D^{*}E^{*}$, and the intersection of $B^{*}C^{*}$ and $AD^{*}$ are collinear, i.e., $M^{*}, N^{*}, O^{*}$ are collinear.

Since $P^{*}$ is the intersection of $M^{*}N^{*}$ and $AD^{*}$, we have $P^{*}=O^{*}$, i.e., $P^{*}$ lies on the line $B^{*}C^{*}$. By the properties of inversion, it follows that in the original figure, $A, B, P, C$ are concyclic, thus $AD \cdot DP = BD \cdot DC \Rightarrow AD = BD = DC = DP = \frac{\sqrt{2}}{2}$, so $AP = \sqrt{2}$.

For rigor, we will prove Pascal's theorem below:

Let $ABCDEF$ be inscribed in a circle (the order of the vertices is irrelevant, i.e., $ABCDEF$ does not need to be a convex hexagon), $AB \cap DE = X, BC \cap EF = Y, CD \cap FA = Z$, then $X, Y, Z$ are collinear.

Proof: Let $AB \cap EF = K, AB \cap CD = M, CD \cap EF = N$. By Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{KX}{XM} \cdot \frac{MD}{DN} \cdot \frac{NE}{EK} = 1, \\
\frac{MZ}{ZN} \cdot \frac{NF}{FK} \cdot \frac{KA}{AM} = 1, \\
\frac{NY}{YK} \cdot \frac{KB}{BM} \cdot \frac{MC}{CN} = 1.
\end{array}
$$

Multiplying the above three equations and using the power of a point theorem, we get
$$
\begin{aligned}
MA \cdot MB & = MD \cdot MC, \\
ND \cdot NC & = NE \cdot NF, \\
KA \cdot KB & = KE \cdot KF,
\end{aligned}
$$

thus
$$
\frac{KX}{XM} \cdot \frac{MZ}{ZN} \cdot \frac{NY}{YK} = 1.
$$

Since $X, Y, Z$ are on $KM, NK, MN$ respectively, by the converse of Menelaus' theorem, $X, Y, Z$ are collinear. Q.E.D.