Solution:

The segments $O B, O C$ are equal, as radii of the circle $^{c}$. Hence $O B C$ is an isosceles triangle and
$$\hat{B}_{1}=\hat{C}_{1}=\hat{x}$$

The chord $B C$ is the bisector of the angle $O \hat{B} A$, and hence
$$\hat{B}_{1}=\hat{B}_{2}=\hat{x}$$
The angles $\hat{B}_{2}$ and $\hat{O}_{1}$ are inscribed to the same arc $O K$ of the circle $^{c_{1}}$ and hence
$$\hat{B}_{2}=\hat{O}_{1}=\hat{x}$$

The segments $K O, K C$ are equal, as radii of the circle $^{c_{2}}$. Hence the triangle $K O C$ is isosceles and so
$$\hat{O}_{2}=\hat{C}_{1}=\hat{x}$$
From equalities $(1),(2),(3)$ we conclude that
$$\hat{O}_{1}=\hat{O}_{2}=\hat{x}$$
and so $O K$ is the bisector, and hence perpendicular bisector of the isosceles triangle $O A C$. The point $K$ is the middle of the arc $O K$ (since $B K$ bisects the angle $O \hat{B} A$). Hence the perpendicular bisector of the chord $A O$ of the circle $^{c_{1}}$ is passing through point $K$. It means that $K$ is the circumcenter of the triangle $O A C$.

From equalities (1),(2),(3) we conclude that $\hat{B}_{2}=\hat{C}_{1}=\hat{x}$ and so $A B \parallel O C \Rightarrow O \hat{A} B=A \hat{O} C$, 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 into the circle $^{c_{2}}$ and correspond to the same arc $O L$. Hence
$$\hat{A}_{1}=\hat{C}_{1}=\hat{x}$$
From (5) and (6) we have $\hat{A}_{1}=\hat{A}_{2}$, i.e. $A L$ is the bisector of the angle $B \hat{A} O$.