Selection Examination

Segments $OB$, $OC$ are equal, as radii of the circle ($c$), and so the triangle $OBC$ is isosceles. Hence: $${\hat{B}}_1={\hat{C}}_1=\hat{x}(1).$$ Figure 3 $BC$ is the bisector of the angle $OBA$, and hence $${\hat{B}}_1={\hat{B}}_2=\hat{x}(2).$$ The angles ${\hat{B}}_2$ and ${\hat{O}}_1$ are inscribed in the same circle and correspond to same arc $OK$ of the circle $(c_1)$. Hence $${\hat{B}}_2={\hat{O}}_1=\hat{x}(3).$$ Similarly we have $KO=KC$ and the triangle $KOC$ is isosceles and hence: $${\hat{O}}_2={\hat{C}}_1=\hat{x}(4).$$ From (1)–(4) we conclude that: ${\hat{O}}_1={\hat{O}}_2=\hat{x}$, i.e. $OK$ is the bisector, hence and perpendicular bisector of the isosceles triangle $OAC$. The point $K$ is the midpoint of the arc $OK$ ($BK$ is the bisector of $OBA$). Hence the perpendicular bisector of the chord $AO$ of the circle ($c_1$) (which is also side of the isosceles triangle $OAC$), is passing through point $K$. It means that $K$ is the circumcenter of the triangle $OAC$.

From (1), (2), we have: ${\hat{B}}_2={\hat{C}}_1=\hat{x}$ and hence $AB\parallel OC$. Therefore $O\hat{A}B=A\hat{O}C$, that is ${\hat{A}}_1+{\hat{A}}_2={\hat{O}}_1+{\hat{O}}_2$ and since ${\hat{O}}_1={\hat{O}}_2=\hat{x}$, we conclude that: $${\hat{A}}_1+{\hat{A}}_2=2{\hat{O}}_1=2\hat{x}.$$ The angles ${\hat{A}}_1$ and ${\hat{C}}_1$ are inscribed in the circle ($c_2$) and correspond to the same arc $OL$. Hence $${\hat{A}}_1={\hat{C}}_1=\hat{x}.$$ From the last two equalities we have that ${\hat{A}}_1={\hat{A}}_2$, that is $AL$ is the bisector of the angle $B\hat{A}O$.