SELECTION EXAMINATION

The triangle $AFK$ is isosceles and hence ${\hat{F}}_1=2{\hat{A}}_1$. The angle ${\hat{A}}_1$ is inscribed in the circle $(c)$ and ${\hat{O}}_1=2{\hat{A}}_1={\hat{F}}_1$, and hence the quadrilateral $BKFO$ is cyclic.

Next we will prove that the quadrilateral $OBK{A}^{\prime }$ is cyclic. In fact, if $S$ be the counter point of $A$ in the circle $(c_1)$. Then the triangle $AKS$ is right angled at $K$, and hence ${\hat{S}}_1=90^{\circ }-{\hat{A}}_1$. The angles ${\hat{S}}_1$ and ${\hat{A}}_1^{\prime }$ are equal (inscribed in the circle $(c_1)$ and they correspond to the same arch $KA$). Hence: ${\hat{A}}_1^{\prime }=90^{\circ }-{\hat{A}}_1$ (1).

From the isosceles triangle $OKB$ we have: ${\hat{B}}_1=90^{\circ }-\frac{{\hat{O}}_1}{2}=90^{\circ }-{\hat{A}}_1$ (2)

From (1), (2) we have: ${\hat{A}}_1^{\prime }={\hat{B}}_1$. Hence the quadrilateral $OBK{A}^{\prime }$ is cyclic.