Denote $(I,r)$ the incircle of $\triangle ABC.$ From $\triangle ABC \sim \triangle AB_1C_1,$ we get $\frac{AO_a}{AI}=\frac{AC_1}{AC}=\cos \widehat{A}=\cos \widehat{T_bAT_c}.$ Since $AI$ is circumdiameter of the A-isosceles $\triangle AT_bT_c,$ then the latter expression means that $O_a$ coincides with the orthocenter of $\triangle AT_bT_c$ $\Longrightarrow$ $T_bO_a \parallel IT_c$ (both perpendicular to AB) and $T_cO_a \parallel IT_b$ (both perpendicular to AC) $\Longrightarrow$ $IT_bO_aT_c$ is a rhombus $\Longrightarrow$ $O_aT_b=O_aT_c=r$ and analogously we'll get $O_bT_c=O_bT_a=O_cT_a=O_cT_b=r$ $\Longrightarrow$ hexagon $T_aO_cT_bO_aT_cO_b$ has equal sides.