SELECTION EXAMINATION

The external angle $\mathrm{E}\hat{Z}A$ of the quadrilateral $AZ\Delta \Gamma$ belongs to the orthogonal triangle $AEZ$, with the acute angle $\mathrm{E}\hat{A}Z=\omega$ equal to the angle $A\hat{B}O$, since $OA=OB$. Hence $\mathrm{E}\hat{Z}A=90^{\circ }-\omega$

Let the extension of the radius $BO$ intersect the circle $c(O,R)$ at $H$. Then $A\hat{\Gamma }H=A\hat{B}H=\omega$ and $$90^{\circ }=B\hat{\Gamma }H=B\hat{\Gamma }A+A\hat{\Gamma }H\Rightarrow B\hat{\Gamma }A=90^{\circ }-A\hat{\Gamma }H=90^{\circ }-\omega .$$ Hence $\mathrm{E}\hat{Z}A=B\hat{\Gamma }A$, and the quadrilateral $AZ\Delta \Gamma$ is cyclic.

Figure 4