Suli's geometry problem

Consider the circles of $ADI$ , $\Gamma$ , $BIC$ . It is well-known that the center of circle $BIC$ is the midpoint of arc $BC$ . Thus, circles $ADI$ and $BIC$ are tangent at $I$ . Because three circles have a common radical center, we can extend $AD, BC,$ and the tangent at $I$ to hit at some point $P$ . Let $F$ denote the foot of the perpendicular from $I$ to $BC$ . By similar triangles, we get $PF \times PE = PI^2 = PD \times PA.$ Thus, by the converse of Power of a Point, $ADFE$ is cyclic. Denote $G$ as the foot of the A-altitiude. We have $\angle IDE = \angle ADE - 90^{\circ} = \angle AFE - 90^{\circ} = \angle FAG$ the last being exterior angle theorem. It is now easy to find that $FG = 2$ and $AG = 15$ to get that $\sin \angle FAG = \frac{2}{\sqrt{229}} \implies \sin^2 \angle FAG = \boxed{\frac{4}{229}}.$