Denote the foot of the altitude from $A$ by $E$ and the midpoint of $OH$ by $L$. Since $\angle ATS = \angle AES = 90^\circ$, the quadrilateral $ATLE$ is cyclic. Thus it is enough to prove that the quadrilateral $ALES$ is cyclic. Let $K$ be the reflection of $H$ about $E$. It is well known that $K$ lies on the circumcircle of $\triangle ABC$. The segment $LE$ is the midline of the triangle $OHK$, so $LE = 0.5OK = 0.5OA = TA$. The segment $TL$ is the midline of the triangle $OAH$, so $TL \parallel AE$. Clearly $TL < AH < AE$, hence $ATLE$ is an isosceles trapezium, which is always cyclic.