# Prove that $A$, $I$ and $H$ are collinear. $$(AC)$$ - 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. But 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.
• This is part of a longer problem so I erased all of the redundancy of the problem and ended up with. $D$ is an introvert but he still wants to be alive so I put him there. – Lê Thành Đạt May 10 at 5:16