Segments $OB$, $OC$ are equal, as radii of the circle ($c$), and so the triangle $OBC$ is isosceles. Hence:
$$\hat{B}_1 = \hat{C}_1 = \hat{x} \qquad (1).$$

Figure 3
$BC$ is the bisector of the angle $OBA$, and hence
$$\hat{B}_1 = \hat{B}_2 = \hat{x} \qquad (2).$$
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
$$\hat{B}_2 = \hat{O}_1 = \hat{x} \qquad (3).$$
Similarly we have $KO = KC$ and the triangle $KOC$ is isosceles and hence:
$$\hat{O}_2 = \hat{C}_1 = \hat{x} \qquad (4).$$
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 $B\hat{A}O$.