Let $\alpha:=\angle BAC$, $\beta:=\angle CBA$, and $\gamma:=\angle ACB$. Without loss of generality, assume that $\beta<\gamma$. Note that $$\angle AHO=\angle AHB+\angle BHO=(\pi-\gamma)+\beta=\pi-(\gamma-\beta).$$
Observe that
$$\angle HOS=2\,\angle HCS=2\,\left(\angle HCB-\angle SCB\right)$$
and
$$\angle HOS=2\,\angle HBS=2\,\left(\angle SBC-\angle HBC\right)\,.$$
Thus,
$$\angle HOS=\left(\angle HCB-\angle SCB\right)+\left(\angle SBC-\angle HBC\right)=\angle HCB-\angle HBC\,,$$
as $\angle SBC=\angle SCB$. That is,
$$\angle HOS=\left(\frac{\pi}{2}-\beta\right)-\left(\frac{\pi}{2}-\gamma\right)=\gamma-\beta\,.$$
Ergo, $$\angle OHS=\frac{\pi-(\gamma-\beta)}{2}=\frac{\angle AHO}{2}\,.$$
In other words, $HS$ is the internal angular bisector of $\angle AHO$.
If $AH=AS$, then $OA\perp HS$. Therefore, the triangle $AHO$ must be isosceles with $AH=HO$ because the internal angular bisector of $\angle AHO$ coincides with the altitude from $H$ to $AO$. In other words, $AH=R$, where $R$ is the radius of the circumcircle $\Gamma$ the triangle $ABC$ (noting that the triangles $ABC$ and $BHC$ have the same circumradius).
Let $D$ be the intersection of the line perpendicular to $BC$ at $B$ and the line perpendicular to $AC$ at $A$. Then, $D$ is on the circle $\Gamma$ and $CD$ is a diameter of $\Gamma$. It is easy to show that $AH=BD$ by noting that $AHBD$ is a parallelogram. Because the right triangle $BCD$ satisfies $\angle CBD=\dfrac{\pi}{2}$ and $$CD=2R=2\,AH=2\,BD\,,$$ we deduce that $$\alpha=\angle BAC=\angle BDC=\dfrac{\pi}{3}\,.$$
In fact, the converse also holds. That is, in an acute triangle $ABC$, $AS=AH$ where $H$ is the orthocenter of the triangle $ABC$ and $S$ is the midpoint of the circular arc $BHC$ if and only if $\angle BAC=\dfrac{\pi}{3}$.