1 (Full score: 50 points) In an acute triangle $A B C$, the altitude $C E$ from $A B$ intersects with the altitude $B D$ from $A C$ at point $H$. The circle with diameter $D E$ intersects $A B$ and $A C$ at points $F$ and $G$, respectively. $F G$ intersects $A H$ at point $K$. Given $B C=25$, $B D=20$, $B E=7$, find the length of $A K$.
From the given, we know $\angle A D B=\angle A E C=90^{\circ}$, so
$\triangle A D B \backsim \triangle A E C$,
$$
\frac{A D}{A E}=\frac{B D}{C E}=\frac{A B}{A C}.
$$
10 points. Since $B C=25, B D=20, B E=7$, it follows that $C D=15, C E=24$. From (1) we have
$$
\left\{\begin{array}{l}
\frac{A D}{A E}=\frac{5}{6}, \\
\frac{A E+7}{A D+15}=\frac{5}{6},
\end{array}\right.
$$

Solving this, we get
$$
\left\{\begin{array}{l}
A D=15 \\
A E=18
\end{array}\right.
$$

Thus, point $D$ is the midpoint of the hypotenuse $A C$ of the right triangle $\triangle A E C$, and $D E=\frac{1}{2} A C=15$.

Connecting $D F$. Since point $F$ lies on the circle with diameter $D E$, we have $\angle D F E=90^{\circ}$, so
$$
A F=\frac{1}{2} A E=9.
$$

Since points $G, F, E, D$ are concyclic, and points $D, E, B, C$ are concyclic, we have
$$
\angle A F G=\angle A D E=\angle A B C,
$$

Therefore, $G F \parallel C B$. Extending $A H$ to intersect $B C$ at $P$, we get
$$
\frac{A K}{A P}=\frac{A F}{A B}.
$$
40 points

Since $H$ is the orthocenter of $\triangle A B C$, we have $A P \perp B C$. Also, since $B A=B C$, we have
$$
A P=C E=24.
$$

From (2), we have
$$
A K=\frac{A F \cdot A P}{A B}=\frac{9 \times 24}{25}=8.64.
$$
50 points