Selection Examination A

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: $${\hat{\Gamma }}_2=\frac{180^{\circ }-{\hat{O}}_1}{2}=\frac{180^{\circ }-2{\hat{E}}_1}{2}=90^{\circ }-{\hat{E}}_1.(2)$$ From relations (1) and (2), we have: ${\hat{\Gamma }}_1={\hat{\Gamma }}_2$. The triangles $\Gamma \Delta E$ and $\Gamma OM$, 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 $\Gamma KO$ are equal and therefore we conclude that $\Gamma EM$ 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 $\Gamma OM$, 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: ${\hat{M}}_1=\hat{E}NZ$ and ${\hat{E}}_2=\hat{M}ZN$. Also, from the isosceles triangle $\Gamma EM$, we have: ${\hat{M}}_1={\hat{E}}_2$. Therefore, we get that: $\hat{E}NZ=\hat{M}ZN$ 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 $\Gamma EZ$, $\Gamma T$ is the bisector of the angle $E\hat{\Gamma }Z$ and $OT$ is the perpendicular bisector of the side EZ. Hence T belongs on the circumcircle of the triangle $\Gamma EZ$.