We know that $\angle ADB = \angle AEC = 90^\circ$, therefore

$$\triangle ADB \sim \triangle AEC,$$
and
$$\frac{AD}{AE} = \frac{BD}{CE} = \frac{AB}{AC} \quad (1)$$
But $BC = 25$, $BD = 20$, and $BE = 7$, so $CD = 15$, and $CE = 24$. From (1), we obtain

Thus, point $D$ is the midpoint of the hypotenuse $AC$ of $\triangle AEC$, and
$$DE = \frac{1}{2}AC = 15.$$

Circle with $DE$ as its diameter, $\angle DFE = 90^\circ$, we have
$$AF = \frac{1}{2}AE = 9.$$
Since four points $G$, $F$, $E$ and $D$ are concyclic, and four points $D$, $E$, $B$ and $C$ are concyclic too, we get
$$\angle AFG = \angle ADE = \angle ABC.$$
Thus $GF \parallel CB$. Extend line $AH$ to intersect $BC$ at point $P$, then
$$\frac{AK}{AP} = \frac{AF}{AB} \qquad (2)$$
Since $H$ is the orthocenter of $\triangle ABC$, $AP \perp BC$. From $BA = BC$ we have
$$AP = CE = 24.$$
Due to (2), we get
$$$$
AK = \frac{AF \cdot AP}{AB} = \frac{9 \times 24}{25} = 8.64.