One, (20 points) The side lengths of squares $A B C D$ and $A E F G$ are $a$ and $b$ respectively, with $a > b$, and $A$ being the common vertex. $D C$ intersects $E F$ at $P$, and $A P \perp F C$.
Find $\angle E A D$.
Let $D P=x, P E=y$. Connect $D E$. Since $\angle D = \angle E=90^{\circ}, D, A, E, P$ are concyclic,
$$
\therefore \angle D A P=\angle D E P.
$$

Extend $A P$ to intersect $F C$ at $H$, then
$$
\begin{array}{l}
\angle A H C=90^{\circ}=\angle D. \\
\therefore \angle D A P=\angle D C H, \\
\angle D E P=\angle D C H, \triangle D P E \sim \triangle F P C, \\
D P: F P=P E: P C,
\end{array}
$$

i.e., $x:(b-y)=y:(a-x)$,
$$
x^{2}-a x=y^{2}-b y.
$$

Using the Pythagorean theorem in $\triangle A D P, \triangle A E P$, we get
$$
x^{2}+a^{2}=A P^{2}=y^{2}+b^{2}.
$$
Subtract (1) from (2) to get $a(a+x)=b(b+y)$.

Let $u=a+x, v=b+y$, then
$$
\begin{array}{l}
a^{2} u^{2}=b^{2} v^{2}, \\
a x=b y+b^{2}-a^{2}.
\end{array}
$$

From (2) and (1), we have
$$
(x+a)^{2}=(y+b)^{2}+2\left(b^{2}-a^{2}\right),
$$

i.e., $u^{2}=v^{2}+2\left(b^{2}-a^{2}\right)$.
From (3) and (5), and using $a>b, b^{2}-a^{2} \neq 0$, it is easy to solve $u=\sqrt{2} b, v=\sqrt{2} a$,
or $x=\sqrt{2} b-a, y=\sqrt{2} a-b$.
Let $\angle E A D=\theta$, then $\angle E P D=180^{\circ}-\theta$. Using the cosine rule in $\triangle A D E, \triangle P D E$, we get
$$
\begin{array}{l}
a^{2}+b^{2}-2 a b \cos \theta=D E^{2} \\
=x^{2}+y^{2}+2 x y \cos \theta, \\
\therefore \cos \theta=\frac{a^{2}+b^{2}-x^{2}-y^{2}}{2(a b+x y)}.
\end{array}
$$

Substituting (6) and simplifying, we get
$$
\cos \theta=\frac{\sqrt{2}}{2}, \theta=45^{\circ}.
$$