Let $D$ be the midpoint of the minor arc $BC$ of $(O)$ and $P$ be the intersection point of the tangents at $B, C$ of $(O)$. We will show that $D$ and $P$ lie on $(AEF)$.

Indeed, we have
$$\angle AEP = \angle AEB = \angle BAC - \angle ABE = \angle ODB - \angle ADB = \angle ADO$$
so $(ADP)$ passes through $E$. Similarly, $(ADP)$ passes through $F$. Now, let $J$ be the center of $(AEF)$, and we will show that $JD \parallel OL$.

From here, since $D$ is the center of circle $(BOC)$ then $JD \perp PX$ and $OX \perp PX$, it follows that $O, X, L$ lie on the same line parallel to $JD$. Since $(ABC)$ intersects $(AEF)$ at $D$ other than $A$, then $\Delta DBE \sim \Delta DCF$ but $DB = DC$ so $DE = DF$. On the other hand, $JE = JF$ since $J$ is the center of $(AEF)$ so $JD \perp EF$. According to the familiar result, $EF \perp OL$, so $JD \parallel OL$ which finishes the proof. $\square$