Consider the linear function $f(X)=\mathbb{P}(X,\Gamma)-\mathbb{P}(X,G)$. Then $$\mathbb{P}(G,\Gamma)=f(G)=\frac{f(E)+f(F)}{2}=\frac{BE^2+BF \cdot CF -EG^2-FG^2}{2}=$$$$\frac{(BM^2+BF\cdot CF+EM^2)-2EG^2}{2}=\frac{(FM^2+EM^2)-2EG^2}{2}=\frac{EF^2-2EG^2}{2}=EG$$Thus inverting about $G$ with radius $EG$ maps $F$, $I$, $H$, and $M$ all to line $BC$, finishing the problem.

Attachments:

$\angle FME = 90, GF = GE$ implies $GF = GM = GE$. Note that $G$ is on the radical axis of the point circle $(E)$ and $(ABC)$ because it lies on the $E$-midline of the triangle $EBC$. This gives $GF^2 = GM^2 = GB \cdot GI = GH \cdot GC$ and because $B,C \in FM$; $F,I,M,H,G$ are cyclic.

We complex bash. Let $\Gamma$ be the unit circle (and let $j$, rather than $i$, denote the coordinate of $I$), so that $$|a|=|b|=|c|=1$$$$e=\frac{2bc}{b+c}$$$$m=\frac{b+c}2$$$$\overline{f} = \frac{b+c-f}{bc}$$$$g = \frac{e+f}2 = \frac{2bc+bf+cf}{2(b+c)}$$$$\overline{g} = \frac{\frac2{bc}+\frac{b+c-f}{b^2c}+\frac{b+c-f}{bc^2}}{2\left(\frac1b+\frac1c\right)} = \frac{b^2+4bc+c^2-bf-cf}{2bc(b+c)}$$$$j = \frac{b-g}{b\overline{g}-1} = \frac{\frac{2b^2-bf-cf}{2(b+c)}}{\frac{b^2+2bc-c^2-bf-cf}{2c(b+c)}} = \frac{c(2b^2-bf-cf)}{b^2+2bc-c^2-bf-cf}$$$$h = \frac{c-g}{c\overline{g}-1} = \frac{\frac{2c^2-bf-cf}{2(b+c)}}{\frac{-b^2+2bc+c^2-bf-cf}{2b(b+c)}} = \frac{b(2c^2-bf-cf)}{-b^2+2bc+c^2-bf-cf}$$$$j-f = \frac{bf^2+cf^2-b^2f-3bcf+2b^2c}{b^2+2bc-c^2-bf-cf} = \frac{(b-f)(2bc-bf-cf)}{b^2+2bc-c^2-bf-cf}$$$$h-f = \frac{bf^2+cf^2-c^2f-3bcf+2bc^2}{-b^2+2bc+c^2-bf-cf} = \frac{(c-f)(2bc-bf-cf)}{-b^2+2bc+c^2-bf-cf}$$$$j-m = \frac{-b^3+b^2c-bc^2+c^3+b^2f-c^2f}{2(b^2+2bc-c^2-bf-cf)} = \frac{(c-b)(b^2+c^2-bf-cf)}{2(b^2+2bc-c^2-bf-cf)}$$$$h-m = \frac{b^3-b^2c+bc^2-c^3-b^2f+c^2f}{2(-b^2+2bc+c^2-bf-cf)} = \frac{(b-c)(b^2+c^2-bf-cf)}{2(-b^2+2bc+c^2-bf-cf)}$$Thus $$\frac{(j-f)(h-m)}{(h-f)(j-m)} = -\frac{b-f}{c-f}$$which is real since $F$ lies on line $BC$. $\blacksquare$

Fix triangle $ABC$ and move the point $F$ along the line $BC$ (Yes, ignore the "which is not on the segment" condition). $degF = 1 \implies degG = 1 \implies degH = degI =1$ Notice that the condition $(IMGF)$ cyclic has degree at most $degGI + degMI + degMF +deg GF= 1 + 1 + 0 + 1= 3$. So $4$ points are enough! $1: F = B$ $2: F = M$ $3: F$ such that $G = M_{bc}$ $4: F$ such that $G = M_a$ Okay! Similarly, we can prove $(HMGF)$ and consequently $(IMHGF)$, as desired.

Let $T$ be the point that satisfies $FBET$ parallelogram. $\angle{TFB}=\angle{EBC}=\angle{BCE},FC||TE\Rightarrow \hspace{1mm}\text{FTEC is cyclic}$ $\angle{TEC}+\angle{FCE}=180^\circ=\angle{TEC}+\angle{TIC}\Rightarrow \text{I lies on FTEC circle}$. So $FB\cdot BC=2FB\cdot BM=IB\cdot BT=2BG\cdot IB\Rightarrow \text{FGMI is cyclic}$. We know that $FG=GM$ so $\angle{FIG}=\angle {GIM}=\angle{FMG}$ so $\triangle {GMB}\sim \triangle {GIM}\Rightarrow GM^2=GB\cdot GI=GH\cdot GC$ so $\angle{GMF}=\angle{MHC}$ and we get $\angle{GIM}+\angle{GHM}=180^\circ$ so $H$ lies on $FGMI$ circle. We're done.

We have $BC$ is polar of $E$ WRT $\Gamma$ so $\bigodot(EF)$ is orthogonal to $\Gamma$. From this, we have $\mathcal{I}^{k = GM^2}_G: \Gamma \longleftrightarrow \Gamma$. Hence $\mathcal{I}^{k = GM^2}_G: F \longleftrightarrow F, M \longleftrightarrow M, B \longleftrightarrow I, C \longleftrightarrow H$. But $B, C, M, F$ are collinear then $G, I, H, F, M$ lie on a circle