1. In the acute triangle $\triangle ABC$, 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 $AK$.
Given that points $B, E, D, C$ are concyclic $\Rightarrow \angle A G F=\angle A E D$ $=\angle A C B \Rightarrow F G \parallel B C$,
Also, $B C=25, B E=7, B D=20, C D=15, C E=24$, by Ptolemy's theorem, $B C \cdot D E+B E \cdot C D=B D \cdot C E$ $\Rightarrow 25 D E+7 \cdot 15=20 \cdot 24 \Rightarrow D E=15$.
Also, $D F \parallel C E, E G \parallel B D \Rightarrow \triangle B C D \sim \triangle D E F, \triangle B C E \sim \triangle D E G$, thus $\frac{D E}{B C}=\frac{D F}{B D}=\frac{E F}{C D}=\frac{D G}{B E}=\frac{E G}{C E}=\frac{3}{5}$, applying Ptolemy's theorem in quadrilateral $D E F G$, we get $F G=\frac{3}{5} B C=9$. And $B F=B E+E F=7+\frac{3}{5} C D=16$, thus $\frac{A F}{A B}=\frac{F G}{B C}=\frac{9}{25} \Rightarrow \frac{A F}{B F}=\frac{9}{16} \Rightarrow A F=9 \Rightarrow A B=25=B C$, therefore the height from $A$ to $B C$ is $A M=C E=24$, so $\frac{A K}{A M}=\frac{9}{25} \Rightarrow A K=\frac{9 \cdot 24}{25}=\frac{216}{25}$.