Let $\overline{AB}$ touch $\omega$ at $D$, $U$ be the midpoint of minor arc $AB$, and $H = \overline{UG} \cap \overline{AF}$. By Pascal on hexagon $BAFCUG$, the three points $D, H, E$ are collinear. Considering the homothety with center $C$ that brings $\omega$ to $\Omega$, we obtain $\overline{E-H-D} \parallel \overline{FU}$. Therefore, $\measuredangle CEH = \measuredangle CFU = \measuredangle CGU = \measuredangle CGH \implies H \in \odot(CGE)$, and $\angle FHE = \angle HFU = \angle AFU = \angle AGU = \angle UGB = \angle HGE \implies AF$ is tangent to $\odot(CGE)$ at $H$.

Let $D=\omega\cap AB,J=DE\cap AF.$ From $\measuredangle CDJ=\measuredangle CDE\stackrel{\text{homothety}}{=}\measuredangle CBF=\measuredangle CAJ$ we obtain that $ADCJ$ is cyclic and therefore $C$ is the Miquel point of $BDJF\implies \measuredangle CJE=\measuredangle CJD=\measuredangle CFB=\measuredangle CGB\implies CJGE$ is cyclic. It's suffice to observe that $\measuredangle ECJ=\measuredangle DCJ+\measuredangle ECD=\measuredangle DAJ+\measuredangle JDA=\measuredangle DJA=\measuredangle EJA.$

Let $CD$ meet $\Omega$ at $T$ and $DE$ meet $AF$ at $H$. Claim $: AHDC$ is cyclic. Proof $:$ Note that $\angle CDH = \angle CDE = \angle CTF = \angle CAH$. Now Note that $\angle CHE = \angle CHD = \angle CAD = \angle CAB = \angle CGB = \angle CGE$ so $CEGH$ is cyclic. Also Note that $\angle CHA = \angle 180 - \angle CDA = \angle 180 - \angle CDE - \angle EDA = \angle 180 - \angle CDE - \angle ECD = \angle 180 - \angle HEC$ so $CEGH$ is tangent to $AF$.