China Mathematical Competition (Extra Test)

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

$$△ADB\sim △AEC,$$ and $$\frac{AD}{AE}=\frac{BD}{CE}=\frac{AB}{AC}(1)$$ 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 $△AEC$, 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 $$\frac{AK}{AP}=\frac{AF}{AB}(2)$$ Since $H$ is the orthocenter of $△ABC$, $AP\perp BC$. 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.