Let the ray $AI$ cuts $\omega$ at $M_A$ and let the exterior angle bisector of the angle $A$ cuts $\omega$ at $N_A$. Let the ray $N_AI$ cuts $\omega$ at $K_A$. Finally, let $I_A$ be the excenter opposite to $A$. We know that $M_AI=M_AB=M_AC=M_AI_A$ (a well-known lemma). Hence, $M_A$ is the center of the circumcircle of $IBC$ and $II_A$ is the diameter of the circumcircle of $IBC$. We have $\angle IAN_A=90^\circ$. Then $\angle M_AK_AI=\angle M_AK_AN_A=\angle M_AAN_A=\angle IAN_A=90^\circ$. We know that $II_A$ is the diameter of the circumcircle of $IBC$. Then, $\angle I_AT_AI=90^\circ$. We have $\angle M_AK_AI=90^\circ=\angle I_AT_AI$. Thus, $I_AT_A//M_AK_A$. Also, $IM_A=M_AI_A$. Then, $K_A$ is the midpoint of the segment $IT_A$. Define $K_B$ and $K_C$ similarly. Look at the homothety with center $I$ and ratio $1/2$. This homothety takes $T_A$ to $K_A$, $T_B$ to $K_B$, $T_C$ to $K_C$. Then, the radius of the circumcircle of $T_AT_BT_C$ is twice the radius of the circumcircle of $K_AK_BK_C$. We know that $K_A, K_B, K_C\in \omega$. Then, the circumcircle of $K_AK_BK_C$ is $\omega$. This completes the proof.

Let $M_A , M_B , M_C$ be the intersections of the exterior bisectors with $\omega$, corresponding to the vertices $A,B,C$ respectively. Let $I_A , I_B , I_C$ be the excenters corresponding to the vertices $A,B,C$ respectively. Since $90 = \angle IAI_B = \angle ICI_B$, $I_B \in (AIC)$. Similarly we get $I_A \in (BIC)$ and $I_C \in (AIB)$. Note that $M_A , M_B , M_C$ are midpoints in $\triangle I_AI_BI_C$. So by a well-known lemma, we get $T_A \in (I_AI_BI_C) , T_B \in (I_AI_BI_C) , T_C\in (I_AI_BI_C)$ and so $I_AI_BI_CT_AT_BT_C$ are concyclic. To finish, just note that $(ABC)$ is the 9-point circle of $\triangle I_AI_BI_C$ and we are done. $\square$

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