SELECTION EXAMINATION

From the inscribed in the circle $c_1$ quadrilateral $A{A}^{\prime }IZ$ we have: $$A{\hat{A}}^{\prime }I=A\hat{Z}I=90^{\circ }=TA{\hat{A}}^{\prime }I.(\alpha )$$ From the inscribed quadrilateral $\Gamma \Delta IE$ (since $\Gamma I$ bisector), we have $${\hat{\Delta }}_1=\frac{\hat{\Gamma }}{2}(1)$$ From the inscribed quadrilateral $B\Delta IZ$ (since $BI$ bisector), we have: $${\hat{\Delta }}_2=\frac{\hat{B}}{2}(2)$$ From the inscribed quadrilateral $B\Delta I{B}^{\prime }$ we have: ${\hat{\Delta }}_3={\hat{B}}_1$. From the inscribed quadrilateral $B{B}^{\prime }{A}^{\prime }A$ we have ${\hat{B}}_1={\hat{A}}_1^{\prime }=90^{\circ }-{\hat{A}}_2^{\prime }$. Hence: ${\hat{\Delta }}_3+{\hat{A}}_2^{\prime }=90^{\circ }$. From the inscribed quadrilateral $AEI{A}^{\prime }$ we have: ${\hat{A}}_3^{\prime }=\frac{\hat{A}}{2}$ Hence: $${\hat{\Delta }}_1+{\hat{\Delta }}_2+{\hat{\Delta }}_3+{\hat{\Delta }}_2^{\prime }+{\hat{\Delta }}_3^{\prime }=180^{\circ }$$ Therefore the quadrilateral $A^{\prime }E\Delta B^{\prime }$ is cyclic. Similarly we prove that the quadrilaterals $\Delta Z{A}^{\prime }{\Gamma }^{\prime }$ and $ZE{\Gamma }^{\prime }{B}^{\prime }$ are cyclic.

Finally, we conclude that the lines $\Delta A^{\prime }$, $E{B}^{\prime }$ and $Z{\Gamma }^{\prime }$ are concurrent at the radical point of three circles.