Problem

Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2015

Tags: geometry, incenter, circumcircle



Let $ABC$ be a triangle with incenter $I$. Line $AI$ intersects circumcircle of $ABC$ in points $A$ and $D$, $(A \neq D)$. Incircle of $ABC$ touches side $BC$ in point $E$ . Line $DE$ intersects circumcircle of $ABC$ in points $D$ and $F$, $(D \neq F)$. Prove that $\angle AFI = 90^{\circ}$