SELECTION EXAMINATION

Since $O_1$ belongs to perpendicular bisector $OM$ of the segment $B\Gamma$ and $AH$, $O{O}_1$, it is enough to prove that $AH=O{O}_1$. Since $AH=2OM$, it is enough to prove that $OM=M{O}_1$. The quadrilateral is cyclic, whereby $\widehat{BH}\Gamma =180^{\circ }-\hat{A}$. Moreover ${\widehat{BO}}_1\Gamma =2\hat{A}$. Therefore the isosceles triangles $BO\Gamma$, $B{O}_1\Gamma$ have all their corresponding angles equal and since they have $B\Gamma$ common side, they are equal. Therefore $B\Gamma$ is the perpendicular bisector of $O{O}_1$, whereby $M$ is the midpoint of $O_{1}O$, and hence $OM=M{O}_1$.