Plain similarity

Let $\angle CBD = \alpha$ . Since triangle $BCD$ is right-angled, we deduce that $CD = 15$ , $\sin\alpha = \tfrac35$ and $\angle BCD = 90^\circ-\alpha$ . Since triangle $CBE$ is right-angled, we deduce that $CE = 24$ and $\sin \angle CBE = \tfrac{24}{25} = 2 \times \tfrac35 \times \tfrac45$ , so that $\angle CBE = 2\alpha$ , and hence $\angle DBE = \alpha$ . Thus $\angle DCE = (90^\circ - \alpha) - (90^\circ - 2\alpha) = \alpha$ Thus $AC = 30$ and $AE = 18$ , and hence $AB = 25$ . Since $\angle BDC = \angle BEC = 90^\circ$ , the circle with $BC$ as diameter passes through both $D$ and $E$ , so that $BCDE$ is cyclic, so $\angle CED = \angle CBD = \alpha$ as well. Thus triangle $CDE$ is isosceles, and $DE = 15$ . Thus $\angle FDE = 2\alpha$ and so, since $DEGF$ is cyclic, $\angle FGA = 2\alpha$ . On the other hand $AK$ passes through the orthocentre $H$ of the triangle $ABC$ , and hence is perpendicular to $BC$ . Thus $\angle GKA = 90^\circ - 2\alpha$ , and hence we deduce that $\angle AKG = 90^\circ$ . But this means that $AK = AG\sin2\alpha = \tfrac{24}{25}AG$ Since $GE = 2 \times \tfrac{15}{2} \times \cos(90^\circ - \alpha) = 9$ , we deduce that $AG = 25 - 7 - 9 = 9$ , so that $AK = \tfrac{216}{25} = \boxed{8.64}$ .