Let $T = \overline{AF} \cap \overline{BE}$. Note that $\triangle IFB \sim \triangle IAE$ since they are isosceles triangles with equal angles, so $I$ is the center of a spiral mapping $FB \mapsto AE$. Therefore, $I$ is the Miquel Point of quadrilateral formed by lines $\{\overline{AF}, \overline{BE}, \overline{AE}, \overline{BF}\ \} \equiv \{\overline{AT}, \overline{BT}, \overline{AG}, \overline{BG}\}$. Simple angle chasing yield $$\measuredangle AIG = \measuredangle AFG = \measuredangle TFB = \measuredangle TIB,$$which means $IG$ and $IT$ are isogonal w.r.t. $\angle AIB$. We just need to prove that $IH$ and $IG$ are isogonal w.r.t. $\angle AIB$ as well. Let $X, Y$ be the midpoints of $AE, BF$ respectively. Observe that $\measuredangle FIA = \measuredangle BIE$ and $\measuredangle YIA = \measuredangle YIB + \measuredangle BIA = \measuredangle AIX + \measuredangle BIA = \measuredangle BIX$. Therefore, $(IY, IX)$, $(IE, IF)$ are pairs of isogonal lines w.r.t. $\angle AIB$. Since $G = \overline{FY} \cap \overline{EX}$ and $H = \overline{FX} \cap \overline{EY}$, therefore isogonal line lemma gives us $IH, IG$ are isogonal w.r.t. $\angle AIB$ as well. So, $I, H, T$ are collinear, i.e. $AF, BE, IH$ are concurrent at $T$.

This is very likely an overcomplicated solution. 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Proof. Let $G'$ be the reflection of $G$ w.r.t. $I$. $I$ is the spiral mapping $EC \mapsto DF$, so $I$ is also the Miquel Point of quadrilateral formed by $\{KD, DE, KC, CF\}$. Then, angle chasing gives us \[ \measuredangle AGB = \measuredangle EGB = \measuredangle EIB = \measuredangle EID = \measuredangle EKD = \measuredangle CKD.\]This means $\measuredangle CKD = \measuredangle AGB = \measuredangle CG'D \implies \triangle CKD$ and $\triangle CG'D$ share the same circumcircle. Therefore, the circumcenters of $\triangle GAB$ and $\triangle KCD$ are symmetrical w.r.t. $I$. $\blacksquare$ Let $O$ be the circumcenter of $\triangle AGB$. We will prove $FN, IO,$ and $EM$ are concurrent. Let $P = \overline{FN} \cap \overline{EM}$. We will prove that $P, O, I$ are collinear. Claim 02 $JI \perp EF$. Proof. We use complex numbers with $I$ as the origin and lowercase complex variables correspond to the point labeled by its uppercase. Therefore, $i = 0$. We pick $a, b$ as arbitrary complex numbers, and \[ f = \frac{b}{r}, e = ar, \]where $r$ is a unitary complex number which rotates $F \mapsto B$ and $A \mapsto E$ with center $I$. It's easy to get $j = (ar^2-b)/(r^2-1)$ (by noting that $(a-j)\cdot r^2 = (b-j)$), and it's easy to check that both $(e-f)/(\overline{e}-\overline{f})$ and $-j/\overline{j}$ both evaluates to $(ar^2-b)/(\overline{a}-\overline{b}r^2)$. Therefore, $EF \perp IJ$. $\blacksquare$. Claim 03 $IJ, IO$ are isogonal w.r.t $\angle AIB$. Proof. Let $A_1$ be the projection of $A$ to $IF$, $B_1$ be the projection of $B$ to $IE$, and $X = \overline{AA_1} \cap \overline{BB_1}$. First, we see that \[ \frac{IA_1}{IB_1} = \frac{IA}{IB} = \frac{IE}{IF}, \]so $F, E, A_1, B_1$ are concyclic. Therefore, \[ \measuredangle IXB_1 = \measuredangle IA_1B_1 = \measuredangle FEI, \]which gives us $IX \perp EF$. By claim 2, we conclude that $I, J, X$ are collinear. Applying Jacobi Theorem to $\triangle ABI$, we obtain $IO, AB_1, BA_1$ are concurrent at a point $Y$. But by isogonal line lemma, since $(IA, IB)$ and $(IA_1, IB_1)$ are pairs of isogonal lines w.r.t. $\angle AIB$, and $X = AA_1 \cap BB_1$, $Y = AB_1 \cap BA_1$, therefore $(IX, IY)$ are also a pair of isogonal lines w.r.t. $\angle AIB$. So, $IJ, IO$ are isogonal w.r.t. $\angle AIB$. $\blacksquare$ Claim 04 $I, J$ are isogonal conjugates w.r.t. $\triangle PFE$ Proof. By claim 2, $FE$ is an altitude of $\triangle FIJ$ erected from $F$. Therefore, by the isogonality of altitude and cevian containing circumcenter, $\measuredangle EFI = \measuredangle JFN = \measuredangle JFP \implies$ $FJ, FI$ are isogonal w.r.t. $\angle EFP$. Similarly, $EJ, EI$ are isogonal w.r.t. $\angle FEP$. Therefore, the claim holds. $\blacksquare$ Let $EE_1$ and $FF_1$ be the diameter of $\odot(EJI)$ and $\odot(FJI)$ respectively. Since $EF$ and $MN$ are both perpendicular to $IJ$, therefore $E_1F_1$ is also perpendicular to $IJ$. Claim 05 $\triangle PF_1J \sim \triangle PIE$. Proof. We know that $EF \parallel MN \parallel E_1F_1$. The angle bisector of $\angle PMJ$ and $\angle PNJ$ meet at $PJ$ since \[ \frac{AN}{NJ} = \frac{AN}{NF_1} = \frac{AM}{ME_1} = \frac{AM}{MJ}. \]But these two lines are the perpendicular bisectors of $JF_1$ and $JE_1$, respectively. Therefore, their meeting point is also the circumcenter of $\triangle JF_1E_1$, and this point lies on $JP$. Since $JI \perp E_1F_1$, this means $JP, JI$ are isogonal w.r.t. $\angle F_1JE_1$. Therefore, \[ \measuredangle F_1JP = \measuredangle IJE_1 = \measuredangle IEE_1 = \measuredangle IEP. \]Combining this with $\measuredangle F_1PJ = \measuredangle IPE$, we obtain the desired triangle similarity result by A.A. $\blacksquare$ Now, we are ready to prove $P, O, I$ are collinear. It's equivalent with proving $\measuredangle(PI, IE) = \measuredangle(OI, IE)$ by angle chasing as follows: \begin{align*} \measuredangle PIE \stackrel{\text{Claim 05}}{=} \measuredangle PF_1J &= \measuredangle FF_1J \\ &= \measuredangle FIJ \\ &= \measuredangle OIE, \end{align*}where the last equality is because $(IJ, IO)$ and $(IF, IE)$ are pairs of isogonal lines w.r.t. $\angle AIB$. $\square$