This is equivalent to showing that $T,I,M$ are collinear, where $M$ is the midpoint of arc $ABC$. Consider an inversion about $B$. This maps $T'$ to the excircle touch-point on $A'C'$, $I'$ to the $B$-excenter of $BA'C'$, and $M'$ to the intersection of the exterior angle bisector of $A'BC'$ and $A'C'$. It is sufficient to show $B,T',I',M'$ are concyclic, but this is clear as $M'BI'=M'T'I'=\pi$.

Dear Mathlinkers, see O.M. Turquie (2003) or http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure I p. 18. Sincerely Jean-Louis

See it's the the general property is mixtilinear in circle that $T,I,M$ are collinear. And then inversion about $B$ gives rest of the thing as done by pi37

@above I think he knows this fact after 3.5 years.