6 TST for BMO

Let $O$ and $I$ be the centers of the circumscribed and inscribed circle of $ABC$, and $A_1$, $B_1$, $C_1$ be the centers of the arcs $\widehat{AB}$, $\widehat{AC}$, $\widehat{BC}$ (not containing the third vertices) of the circumscribed circle. By symmetry with respect to $A{A}_1$ we have $A_1{C}_a=A_{1}B=A_{1}C$ and together with $O_{c}C=O_c{C}_a$ it follows that $A_1{O}_c$ is the segment bisector of $C{C}_a$. Analogously, $B_1{O}_c$ is the segment bisector of $C{C}_b$. Thus, the quadrilateral $O{A}_1{O}_c{B}_1$ is a parallelogram with $O{A}_1=O{B}_1$, i.e. rhombus. In other words, $O$ and $O_c$ are symmetric about $A_1{B}_1$, and this is also true for $C$ and $I$, i.e. $C{O}_c$ is symmetric to $OI$ with respect to $A_1{B}_1$. Analogously, $A{O}_a$ and $B{O}_b$ are symmetric on $OI$ with respect to $B_1{C}_1$ and $A_1{C}_1$, respectively. Since $I$ is the orthocenter of $A_1{B}_1{C}_1$, finally the lines $A{O}_a$, $B{O}_b$, and $C{O}_c$ intersect at the Anti-Steiner point for the line $OI$ and triangle $A_1{B}_1{C}_1$ (circumscribed about $ABC$). $\square$