KMO 2019 Problem 6

In acute triangle $ABC$ , $AB>AC$ . Let $I$ the incenter, $\Omega$ the circumcircle of triangle $ABC$ , and $D$ the foot of perpendicular from $A$ to $BC$ . $AI$ intersects $\Omega$ at point $M(\neq A)$ , and the line which passes $M$ and perpendicular to $AM$ intersects $AD$ at point $E$ . Now let $F$ the foot of perpendicular from $I$ to $AD$ . Prove that $ID\cdot AM=IE\cdot AF$ .