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 $AA_1$ we have $A_1C_a = A_1B = A_1C$ and together with $O_cC = O_cC_a$ it follows that $A_1O_c$ is the segment bisector of $CC_a$. Analogously, $B_1O_c$ is the segment bisector of $CC_b$. Thus, the quadrilateral $OA_1O_cB_1$ is a parallelogram with $OA_1 = OB_1$, i.e. rhombus. In other words, $O$ and $O_c$ are symmetric about $A_1B_1$, and this is also true for $C$ and $I$, i.e. $CO_c$ is symmetric to $OI$ with respect to $A_1B_1$. Analogously, $AO_a$ and $BO_b$ are symmetric on $OI$ with respect to $B_1C_1$ and $A_1C_1$, respectively. Since $I$ is the orthocenter of $A_1B_1C_1$, finally the lines $AO_a$, $BO_b$, and $CO_c$ intersect at the Anti-Steiner point for the line $OI$ and triangle $A_1B_1C_1$ (circumscribed about $ABC$). $\square$