Let $X=AD \cap BE$, $K=AH \cap (ACDF)$ and $L=BG \cap (BCEF)$. From the cyclic quadrilaterals we get that: $\angle HKF = \angle AKF = \angle ACF = \angle ECF = \angle EBF = \angle HBF => HKBF$ is cyclic. Hence we get that $\angle XBK = \angle HBK = \angle HFK = \angle DFK = \angle DAK = \angle XAK$, and so $BKXA$ is cyclic. Analogously, $ALXB$, is cyclic and hence $A,L,X,K,B$ are all concyclic. Now from radical axis on $(CDKFA),(CELFB)$ and $(ALXKB)$ we get that $CF,AK,BL$ are concurrent and hence $CF,AH,BG$ are concurrent

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Here's a solution with Ceva's theorem We just need to prove that $\frac{\sin \angle ACF}{\sin \angle BCF}\cdot\frac{\sin \angle CBG}{\sin \angle ABG}\cdot\frac{\sin \angle BAH}{\sin \angle CAH}=1$. Notice that $\angle EBF=\angle ACF=\angle ADF$, $\angle BEF=\angle BCF=\angle DAF$, hence $\triangle BEF\sim \triangle DAF$, then $\triangle BDF\sim \triangle EAF$. Then, $$\frac{\sin \angle ACF}{\sin \angle BCF}=\frac{\sin \angle ADF}{\sin \angle DAF}=\frac{AF}{DF}=\frac{AE}{BD}$$$$\frac{\sin \angle CBG}{\sin \angle ABG}=\frac{DG}{AG}\cdot\frac{AB}{BD},\frac{\sin \angle BAH}{\sin \angle CAH}=\frac{BH}{HE}\cdot\frac{AE}{AB}$$Hence, $$\frac{\sin \angle ACF}{\sin \angle BCF}\cdot\frac{\sin \angle CBG}{\sin \angle ABG}\cdot\frac{\sin \angle BAH}{\sin \angle CAH}=\frac{AE}{BD}\cdot\frac{DG}{AG}\cdot\frac{AB}{BD}\cdot\frac{BH}{HE}\cdot\frac{AE}{AB}=\frac{AE^2}{BD^2}\cdot\frac{DG}{AG}\cdot\frac{BH}{HE}=1$$In the last step we used that $$\frac{AE}{BD}\cdot\frac{DG}{AG}=\frac{\sin \angle DFG}{\sin \angle AFG}=\frac{\sin \angle EFH}{\sin \angle BFH}=\frac{BD}{AE}\cdot \frac{HE}{BH}$$Hence we are done. $\square$

Also given at North Macedonia Balkan MO 2024 TST and some revenge ELMOs at MOP.

As some students found, just one Pappus finishes this problem without using the definition of point F(and my solution was like #3).

Indeed, the circles $ACDF$ and $BCEF$ are totally not needed, the point $F$ could be as arbitrary as we wish. Applying Pappus' theorem to the lines $ADG$ and $BEH$ gives that $AE \cap BD = C$, $DH \cap EG = F$ and $AH \cap BG$ are collinear, done. :X Good exercise for Pappus, but I am very glad this one was not selected for JBMO (not that it had much of a chance, as it involved radical axis), otherwise it would have been shameful for the competition.