Let $\hat{E}_i = \Gamma \hat{E} Z$. Then from the orthogonal triangle $\Gamma \Delta E : \hat{\Gamma}_1 = 90^\circ - \hat{E}_1$ (1)
Since $\Gamma \hat{O} Z = \hat{O}_1$ is the corresponding central angle of $\hat{E}_1 = \Gamma \hat{E} Z$, then
$\hat{O}_1 = 2\hat{E}_1$ and from the isosceles triangle $\Gamma OZ$ we have:
$$\hat{\Gamma}_2 = \frac{180^\circ - \hat{O}_1}{2} = \frac{180^\circ - 2\hat{E}_1}{2} = 90^\circ - \hat{E}_1. \quad (2)$$
From relations (1) and (2), we have: $\hat{\Gamma}_1 = \hat{\Gamma}_2$.

The triangles $\Gamma \Delta E$ and $\Gamma O M$, are equal [$\Gamma \hat{\Delta} E = \Gamma \hat{O} M = 90^\circ$, $\hat{\Gamma}_1 = \hat{\Gamma}_2$ and $\Gamma \Delta = \Gamma O$]. Also the triangles $\Gamma K \Delta$ and $\Gamma K O$ are equal and therefore we conclude that $\Gamma E M$ is isosceles triangle and that $\Gamma K$ is the bisector of the angle $\hat{E} \Gamma M$ and perpendicular bisector of the segment $EM$. (A)
From the equality of triangles $\Gamma \Delta E$ and $\Gamma O M$, we conclude that: $\Gamma \hat{E} Z = \Gamma \hat{M} O$, and therefore $N \hat{E} Z = N \hat{M} Z$. Hence the quadrilateral EMZN is cyclic and we have the following equalities: $\hat{M}_1 = \hat{E} N Z$ and $\hat{E}_2 = \hat{M} Z N$.
Also, from the isosceles triangle $\Gamma E M$, we have: $\hat{M}_1 = \hat{E}_2$.
Therefore, we get that: $\hat{E} N Z = \hat{M} Z N$ and then EMZN is isosceles trapezium.

The center of the circumcircle of EMZN is the point of intersection of two perpendicular bisectors of its sides. Let T is the point of intersection of $\Gamma K$ (perpendicular bisector of EM) and the perpendicular bisector of EZ. In the triangle $\Gamma E Z$, $\Gamma T$ is the bisector of the angle $E \hat{\Gamma} Z$ and $O T$ is the perpendicular bisector of the side EZ. Hence T belongs on the circumcircle of the triangle $\Gamma E Z$.