First, we prove the following lemma.

Lemma. Let $I$ and $O$ be the incenter and circumcenter of $\triangle ABC$, respectively. Let the line through $I$ perpendicular to $IO$ meet $BC$ and the external bisector of $\angle BAC$ at $X$ and $Y$, respectively. Then $IY = 2IX$.

Proof of Lemma. Let $I_a, I_b, I_c$ be the excenters of $\triangle ABC$ opposite to the vertices $A, B, C$, respectively.
Apply the homothety centered at $I$ with ratio $2$, taking $A, B, C, X, O$ to $A', B', C', X', O'$, respectively. Since $I$ is the orthocenter of $\triangle I_aI_bI_c$ and $O$ is the nine-point center of $\triangle I_aI_bI_c$, $O'$ is the circumcenter of $\triangle I_aI_bI_c$. But $O'$ is also the circumcenter of $\triangle A'B'C'$, and their circumradii are equal (both being twice the circumradius of $\triangle ABC$), so the six points $A', B', C', I_a, I_b, I_c$ are concyclic.

Since $X$ lies on $BC$, $X'$ lies on $B'C'$. Also it is clear that $B', B, I, I_b$ are collinear, $C', C, I, I_c$ are collinear, and $O', O, I$ are collinear.

Consider the quadrilateral $I_bB'C'I_c$; by the butterfly theorem, $IY = IX'$.
Also $IX' = 2IX$, so $IY = 2IX$, which proves the lemma.

Back to the original problem. Let $P$ be the intersection of the external bisector of $\angle BAC$ with $IY$, let $Q$ be the intersection of $IY$ with $BC$, and let $S$ and $R$ be the intersections of $AI$ with $BC$ and $L$, respectively. By the lemma, $IP = 2IQ$, but clearly $IY = IQ$, so $Y$ is the midpoint of $IP$.

Since $IP$ is the hypotenuse of the right triangle $PAI$, we have $IY = AY$, i.e., $\triangle YAI$ is isosceles, so $\angle YAI = \angle YIA$.

Therefore
$$\begin{align*} \angle YAO &= \angle YAI - \angle IAO = \angle XIA - (90^\circ - \angle C - \frac{\angle A}{2}) \\ &= \angle C + \frac{\angle A}{2} - \angle AIX = \angle ASB - \angle AIX \\ &= \angle YRI - \angle AIX = \angle YXO, \end{align*}$$
which shows that $A, X, O, Y$ are concyclic. This completes the proof.