We will prove that the circle $c_3$ passes from the center $O$ of $c(O,R)$.
From the inscribed quadrilateral $O\Delta\Gamma E$ (in the circle $c_3$) we get: $\hat{O}_1 = \hat{\Gamma}$.
From the inscribed quadrilateral $O\Delta BZ$ (in the circle $c_1$) we get: $\hat{O}_2 = \hat{B}$.
Summing up the above equations we find:
$\hat{O}_1 + \hat{O}_2 = \hat{B} + \hat{\Gamma} \Rightarrow EOZ = \hat{B} + \hat{\Gamma} = 180^\circ - \hat{A}$,
and therefore the quadrilateral $AEOZ$ is cyclic.

Now we are going to prove that the circles $c_1, c_2, c_3$ are equal.
From the cyclic quadrilateral $O\Delta BZ$ we have: $\hat{\Delta}_2 = \hat{Z}_2$.

Figure 1
From the cyclic quadrilateral $O\Delta\Gamma E$ we have: $\hat{\Delta}_2 = \hat{E}_1$ and hence:
$$\hat{\Delta}_2 = \hat{Z}_2 = \hat{E}_1.$$
These three angles go in the equal chords $OB$, $O\Gamma$ and $OA$ of the circles $c_1, c_2$ and $c_3$, respectively. Therefore these three circles are equal.

Now in the equal circles $c_1$ and $c_2$, the angles $\hat{Z}_1$ and $\hat{\Delta}_1$ go in the equal chords $OK$ and $OM$ ($OK = OM = R$), and so: $\hat{Z}_1 = \hat{\Delta}_1$.
From the last equality we conclude that the points $K$, $\Delta$, $M$ are collinear. Similarly we prove that the points $M$, $E$, $N$ and $N$, $Z$, $K$ are collinear.

From the equalities of angles $B\hat{\Delta}K = \Gamma\hat{\Delta}M$ and $\Gamma\hat{E}M = A\hat{E}N$ we get the equality of segments $AN = BK = \Gamma M$ (chords of the circle $c(O,R)$).

The triangles $AB\Gamma$ and $KMN$ have common circumcenter $O$ and the triangle $KMN$ is the image of $AB\Gamma$ in the rotation $R(O, \omega)$, where
$$A\hat{O}N = B\hat{O}K = \Gamma\hat{O}M = \hat{\omega}. \text{ Hence the triangles are equal.}$$