$(AC)$ - Circle diameter $AC$ - is the circumcircle of cyclic quadrilateral $ABCD$. $F$ is a point on the smaller arc of $CD$ such that $FC < FD$. $BF$ intersects $CD$ at $N$. $M$ is a point on line $CF$ such that $MN \perp AC$ and intersects $BC$ at $I$. $AM \cap (AC) = G$, $GN \cap (AC) = H$ ($H \not\equiv G \not\equiv A$). Prove that $A$, $I$ and $H$ are collinear.
I am trying to prove that $\widehat{IHN} = \widehat{AHG}$ but I don't know how. (I wonder if I am a normal human being if I can't solve this, at least an average student could.)