Denote in the triangle $ABC$ by $T_A,T_B,T_C$ the touch points of the exscribed circles of $\vartriangle ABC$, tangent to sides $BC, AC$ and $AB$ respectively. Let $O$ be the center of the circumcircle of $\vartriangle ABC$, and $I$ is the center of it's inscribed circle. It is known that $OI\parallel AC$. Prove that $\angle T_A T_B T_C= 90^o - \frac12 \angle ABC$. (Anton Trygub)