Intersecting Circumcircles I came across this interesting problem which I have tried to solve for many days.Consider a scalene triangle ABC. The Euler Line and circumcircle are drawn. G is a point on the Euler Line and F is a point on the circumcircle. The intersection of AF and BG is H. The circumcircle of FGH is drawn. It intersects the previous circumcircle at I. Prove that I is independent of F.
Here is the Geogebra File.

 A: It does not matter that $G$ lies on the Euler line.
We can consider $\phi:F\to H$ as a projectivity between the circumcircle $\Gamma$ of $ABC$ and the line $BG$. Let $I'$ be the intersection between $\Gamma$ and the circumcircle of $F'H'G$, where $H'=\phi(F')$, and $I''$ be the intersection between $\Gamma$ and the circumcircle of $F''H''G$, where $H''=\phi(F'')$. 

Let $J$ be the intersection between $BG$ and $\Gamma$. We have:
$$\widehat{GI'J}=\widehat{GI'F'}+\widehat{F'I'J}=\widehat{AH'B}+\widehat{F'BH'}=\pi-\widehat{BF'H'}=\widehat{AF'B},$$
$$\widehat{GI''J}=\widehat{GI''F''}+\widehat{F''I''J}=\widehat{AH''B}+\widehat{F''BH''}=\pi-\widehat{BF''H''}=\widehat{AF''B},$$
and since $\widehat{AF'B}=\widehat{AF''B}$, it follows that $\widehat{GI'J}=\widehat{GI''J}$, from which $I'\equiv I''$, as wanted.
With the same trick of splitting an angle in many parts, then consider equivalent angles on different circles, we can prove that the intersection $K$ of $AC$ and $BG$ lies on the circumcircle of $GCI$. This gives another proof of the uniqueness of $I$, as intersection between the circumcircle of $ABC$ and the circumcircle of $GKC$.
