$(AC)$ - 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.)
You can notice that the lines $AF$ and $CG$ intersect at the orthocentre of $ACM$ and therefore at the line $NM$ so you can apply Pascal's theorem to the hexagon $AHGCBF$ and note that it implies that the intersection of $AH$ and $BC$ lies on $NM$ which yields the result.
I am not sure whether it is the kind of solution you are looking for. Also do note that point $D$ doesn't play any role in this picture - it's useless both in defining the problem as well as solving it.