Question 122
Let the circumcircle of triangle $ABC$ be $\Gamma$ and its incentre $I$. Let $M$ be the midpoint of side $BC$. Drop the perpendicular from $I$ to $BC$, with foot $D$. The line through $I$ perpendicular to $AI$ meets $AB$, $AC$ at $F$, $E$ respectively. Let the circumcircle of triangle $AEF$ meet $\Gamma$ again at $X$. Prove that the intersection point of the lines $XD$ and $AM$ lies on $\Gamma$.