Consider a point $G$ in the interior of a parallelogram $ABCD$. A circle $\Gamma$ through $A$ and $G$ intersects the sides $AB$ and $AD$ for the second time at the points $E$ and $F$ respectively. The line $FG$ extended intersects the side $BC$ at $H$ and the line $EG$ extended intersects the side $CD$ at $I$. The circumcircle of triangle $HGI$ intersects the circle $\Gamma$ for the second time at $M \ne G$. Prove that $M$ lies on the diagonal $AC$.