Let $M$ - midpoint of the arc $BC$. Inversion $(M;MC)$ preserves $(AOL)$>

, hence $O'$ - image of $O$ is on $(AOL)$ and fixed. Center of $(AOL)$ is on perp bisector of $OO'$ which is parallel to $BC$.

Let the circumcircle of triangle $AOL$ intersect the line $BC$ again $D$. Wlog, let $\angle ACB> \angle CBA$. $\angle AOD=\angle ALD=\dfrac{\angle BAC}{2}+\angle CBA$ and $\angle OAD=\angle CAO-\angle CAD=(90^\circ-\angle CBA)- \dfrac{\angle BAC}{2}=90^\circ-\angle AOD$ then $\angle OAL+\angle AOD=90^\circ$ $\Rightarrow OD \perp AL$ Let the center of the circumcircle of triangle $AOL$ be $O'$, the projections of the point $O'$ on the lines $AO$ and $BC$ be $M$ and $N$, respectively. Since the diagonals of the $AOLD$ are perpendicular, $\widehat{OL} +\widehat{AD}=180^\circ \Rightarrow$ $\angle AO'O+ \angle LO'D=180^\circ \Rightarrow \angle AO'M+ \angle LO'N=90^\circ$. Moreover, since $AO'=O'L$, right triangles $AO'M$ and $O'LN$ are congruent then $AM=O'N$ . Consequently the length $O'N$ is fixed.