Bariz - Problem

Let $X$ be the mid-point of $AO$ and $D$ be the foot of the perpendicular of $A$ on $BC$.The following claim is the crux of the problem. Claim: $A,X,L,D,S$ all lie on a circle. proof.Firstly,as $\angle AXS=\angle ADS=90^{\circ}$ hence $AXDS$ is cyclic quadrilateral. Now,Let $M$ be the mid-point of $ BC$.So the reflection of $M$ on $L$ lies on $AH$ as well as on nine-point circle of $ABC$.Call this point $T$.Also observe that $MT||AO$.Also if $G$ be the midpoint of $DM$ then $X,L,G$ are collinear. Hence, $\angle DLG=\frac{1}{2}\angle DLG=\angle DTM=\angle DAO=\angle DAX$. So $DAXL$ lies on a circle.$\blacksquare$. Finally observe that as $XL$ is the $O$-midline of $OAH$ so $XL||AD$. So $SL$ and $SX$ are isogonal w.r.t $\angle DOA$. So, $\angle ASL=\angle XSD=\angle DAX=\angle HAO$. We are done.