In the acute triangle $ABC$, the altitude $CE$ from $AB$ intersects with the altitude $BD$ from $AC$ 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 $BC=25, BD=20, BE=7$. Find the length of $AK$.
8. 64
I. 【Analysis and Solution】From the given conditions, we know $\angle A D B=\angle A E C=90^{\circ}$,
$\therefore \triangle A D B \sim \triangle A E C \Rightarrow \frac{A D}{A E}=\frac{B D}{C E}=\frac{A B}{A C}$
Since $B C=25, B D=20, B E=7 \Rightarrow C D=15, C E=24$.
From (1) we have $\left\{\begin{array}{l}\frac{A D}{A E}=\frac{5}{6} \\ \frac{A E+7}{A D+15}=\frac{5}{6}\end{array} \Rightarrow\left\{\begin{array}{l}A D=15 \\ A E=18\end{array}\right.\right.$.
$\therefore$ Point $D$ is the midpoint of the hypotenuse $A C$ of the right triangle $\triangle A E C$, $\Rightarrow D E=\frac{1}{2} A C=15$.
Connecting $D F$, since point $F$ lies on the circle with $D E$ as its diameter, $\therefore \angle D E F=90^{\circ}$.
$\therefore A F=\frac{1}{2} A E=9$.
$\because G, F, E, D$ are concyclic, and $D, E, B, C$ are concyclic,
$\therefore \angle A F G=\angle A D E=\angle A B C$.
$\therefore G F \parallel C B$, extending $A H$ to intersect $B C$ at $P$, we have $\frac{A K}{A P}=\frac{A F}{A B}$
$\because H$ is the orthocenter of $\triangle A B C$, $\therefore A P \perp B C$.
Also, $\because B A=B C, \therefore A P=C H=24$.
$\therefore$ From (2) we have $A K=\frac{A F \cdot A P}{A B}=\frac{9 \times 24}{25} \approx 8.64$