One, (50 points) As shown in Figure 4, in the acute triangle $\triangle ABC$, the altitude $CE$ from $A B$ intersects with the altitude $BD$ from $A C$ at point $H$. The circle with diameter $DE$ intersects $AB$ and $AC$ at points $F$ and $G$, respectively. $FG$ intersects $AH$ at point $K$. Given that $BC=25, BD=20, BE=7$. Find the length of $AK$.
Given that $\angle A D B=\angle A E C=90^{\circ}$. Therefore,
$\triangle A D B \backsim \triangle A E C$.
Thus, $\frac{A D}{A E}=\frac{B D}{C E}=\frac{A B}{A C}$.
From the given, $C D=15, C E=24$. From (1) we have
$\left\{\begin{array}{l}\frac{A D}{A E}=\frac{5}{6}, \\ \frac{A E+7}{A D+15}=\frac{5}{6}\end{array} \Rightarrow\left\{\begin{array}{l}A D=15, \\ A E=18 .\end{array}\right.\right.$
Thus, point $D$ is the midpoint of the hypotenuse $A C$ of the right triangle $\triangle A E C$, so $D E=\frac{1}{2} A C=15$.
Connecting $D F$. Since point $F$ lies on the circle with diameter $D E$, then $\angle D F E=90^{\circ}, A F=\frac{1}{2} A E=9$.
Since $G, F, E, D$ and $D, E, B, C$ are concyclic, then $\angle A F G=\angle A D E=\angle A B C, G F / / C B$.
Extending $A H$ to intersect $B C$ at $P$, then
$$
\frac{A K}{A P}=\frac{A F}{A B} \text {. }
$$

Since $H$ is the orthocenter of $\triangle A B C$, then $A P \perp B C$. Also, since $B A=B C$, we have $A P=C E=24$. From (2) we have
$$
A K=\frac{A F \cdot A P}{A B}=\frac{9 \times 24}{25}=8.64
$$