Consider the inversion of pole $C$ and power $CG^2=CF^2$. It fixes $\zeta_1$, and it turns $\zeta_2$ into a line which must be tangent to $\zeta_1$ in the point which is the intersection between $\zeta_1$ and $AC$. This point is $B$, so $\zeta_2$ is turned into the line $DE$. This means that the points $D,E$ are fixed by the inversion (since they lie both on $\zeta_2$ and on the image of $\zeta_2$), so $CD=CE=CG=CF$, meaning that $E,D,F,G$ all lie on a circle centered at $C$.

Consider the homotety of center $A$ which tranforms ${\zeta_1}$ to $\zeta_2$ then $DE$ will transform to the tangent of ${\zeta_2}$ at $C$ parallel to $DE$. Hence $CD=CE$ and $\angle{CEB}=\angle{CAE}$ and from here $CD^2=CE^2=CB\cdot{CA}=CG^2=CF^2$ and we are done.

I was in the contest. I'm not that good at geometric transformation, so I straightforwardly calculated the distances from C to those points, and I found them equal. Sorry for my ignorance concerning Latex!