I drew a wrong diagram, sorry

$AM$ and $FO$ meet at $H$ ,and $(KBG)$ and AB meet at $I$. $\angle{FOC} = \angle{FCG} = \angle{CAH}$ so $C,O,A,H$ are concyclic. $\angle{OHA} = \angle{OCA} = \angle{OAC}$ so $MO^2 = AO^2 = FO \times HO$ by a similarity. so $\angle{OMF} = \angle{OHM} = \angle{OAF}$ so $O,F,M,A$ are also concyclic. $\angle{FOC} = \angle{FCB} = \angle{MAF} = \angle{MOF}$ so $\triangle{MOF} \equiv \triangle{COF}$ by SAS and $\triangle{MOG} \equiv \triangle{COG}$ by SAS too. So $\angle{KGF} = \angle{KCF}$ and $K,G,C,F$ are concyclic. Then $\angle{FMO} = \angle{FGC} = \angle{FKO}$ and $\angle{KMA} = \angle{KGB} = \angle{KIB}$ so $A,M,F,O,K,I$ are concyclic. Hence we are done.

We have that: $\angle AFO= \angle FCO + \angle COF = \angle ACO + \angle ACB = \angle CAO + \angle CAM = \angle OAM = \angle OMA$, hence $AMFO$ is cyclic and so: $\angle MOF = \angle MAF = \angle MAC = \angle ACB = \angle COF$. Using this angle equality and that $OC=OM$, we conclude that $FO$ is the perpendicular bisector of $MC$. Hence: $\angle KGF = \angle MGF = \angle CGF = \angle OFA - \angle GCF= \angle OMA - \angle ACB = \angle OAM - \angle ACB =\angle OAM - \angle CAM = \angle OAC = \angle OCA = \angle KCF$ Fom the above angle equality we conclude that $GKFC$ is cyclic. So we have that: $\angle MKF = \angle FCG = \angle ACB = \angle MAF$, so $AMFK$ is cyclic. Hence, $A,K,O,F,M$ are all concyclic. Now let $(BGK) \cap AB = N$. Using the cyclic quadrilaterals and that $AM \parallel BC$ we get that: $\angle BNK = \angle KGC = \angle MGC = \angle AMG = \angle AMK = \angle AOK$, and so $N$ lies on $(AKO)$, as needed.

Attachments:

Notice that \[\angle FCO = 180^{\circ} - \angle OFC - \angle COF = 180^{\circ} - \angle OFC - \angle ACB = \angle FGC. \]Since $OA = OM$ and $AM \parallel BC$, \[\angle AMO = \angle MAO = \angle MAC + \angle CAO = \angle ACB + \angle ACO = \angle COF + \angle ACO = \angle AFO \]therefore $AMFO$ is cyclic. Now, $\angle FOM = \angle FAM = \angle ACB = \angle COF$ and as $OM = OC$, $\Rightarrow GOF$ is perpendicular bisector of $MC$ $\Rightarrow \angle KGF = \angle FGC = \angle FCO$ $\Rightarrow KGCF$ is cyclic. $\Rightarrow \angle MKO = 180^{\circ} - \angle GKC = 180^{\circ} - \angle GFC = \angle AFO = \angle MAO.$ $\Rightarrow K$ lie on $\odot (AMFO).$ Finally, let $T = \odot (BGK) \cap AB \neq B$, \[\angle BTK = \angle KGC = \angle AFK\]thus $T$ lie on $\odot (AOK)$ $\blacksquare$

We start off by proving the following claims. Claim : Points $A$ , $M$ , $F$ and $O$ lie on the same circle (say $\gamma$). Proof : Note that, \[\measuredangle MOF = \measuredangle MOC + \measuredangle COF = 2\measuredangle MAC + \measuredangle ACB = 2\measuredangle MAC + \measuredangle CAM = \measuredangle MAC\]which proves the claim. One thing to note from this is that $\measuredangle MOF = \measuredangle FOC$ which implies that $OF$ is the $\angle MOC$-bisector. Further, since $OM=OC$, it follows that $OF$ is in fact also the perpendicular bisector of $MC$. Now, as a result of this, $\measuredangle MGO = \measuredangle OGC$ as well. Using this we can prove the following claim. Claim : $K$ also lies on $\gamma$. Proof : Note that by the above established angle equality we have, \[\measuredangle KMA = \measuredangle GMA = \measuredangle MGC = 2\measuredangle OGC = 2\measuredangle OCA = \measuredangle KOA \]from which it is clear that $K$ also lies on $\gamma$ as claimed. Now, we are almost there. Let $X = \overline{AB} \cap \gamma$. Then, \[\measuredangle KXB = \measuredangle KOA = 2\measuredangle OCA = 2\measuredangle OGC = \measuredangle MGC = \measuredangle KGB\]from which it follows that $X$ also lies on $(KGB)$. Thus, indeed the circumcircles of the triangles $BGK$ and $AOK$ meet on $AB$.