Hint:

$ABH$ is an equilateral triangle, $C$ is its center, so $H$ is the orthocenter of $ABC$.

So the smallest angle of $ABC$ is $30°$.

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 $\angle BAC=\dfrac{\pi}{3}$.

## Best Answer

We use this fact that the bisector of $\angle BFC$ and Perpendicular bisector of BC meet on circumcircle of triangle BFC and the bisector of $\angle BHC$ and Perpendicular bisector of BC meet on circumcircle of triangle BHC. These two circle have common chord BC and the meeting points are both on perpendicular bisector of BC, that means bisectors of angles $\hat {BFC}$ and $\hat{BHC}$meet at G which is on BC.