Iranian Mathematical Olympiad

Throughout the solution, let $\omega$ and $\omega_a$ be the circumcircle of triangle $ABC$ and the excircle of $ABC$ tangent to side $BC$, respectively. Furthermore, $O$ and $I_a$ are the centers of $\omega$ and $\omega_a$, respectively.
Under an inversion with center $A$ and power $AB \times AC$, and then a reflection with respect to the bisector of $\angle BAC$, $B'$ is sent to some point of line $AB$, and $C'$ to some point of line $AC$. Denote by $B''$ and $C''$ these two points, respectively. Note that
$$AB' \cdot AC'' = AC' \cdot AB'' = AB \cdot AC = AI \cdot AI_a.$$
Therefore, the quadrilaterals $II_aB''C'$ and $II_aC''B'$ are cyclic. Since $\angle I_aIB' = 90^\circ$, we get $I_aC'' \perp AC$ and hence $C''$ is the tangency point of $\omega_a$ and $AC$. Similarly, $B''$ is the tangency point of $\omega_a$ and $AB$.
Now, we are going to find the image of $B_1$ and $C_1$ under the transformation mentioned above. Denote by $M$ and $N$ the midpoints of arcs $ACB$ and $ABC$ of circle $\omega$, respectively. Then $\angle BAC_1 = \angle CAM$. We know that under the transformation, line $BC$ transforms to circle $\omega$. Hence $C_1$ is sent to $M$. Analogously, $B_1$ is sent to $N$. So $T$ is sent to the intersection point of lines $NC''$ and $MB''$, say $T'$. Note that the tangent to $\omega$ at $M$ is parallel to the tangent to $\omega_a$ at $C''$. Thus the direct homothetic center of $\omega$ and $\omega_a$, lies on line $NC''$. Denote this point by $T''$. By a similar argument, $T''$ lies on $MB''$ and hence $T'' = T'$ lies on line $OI_a$. This implies that $T'I_a$ is perpendicular to $\omega$. Since the transformation above preserves angles, we deduce that the circumcircle of $ATI$ (the transformation of $T'I_a$) is perpendicular to $BC$ (the transformation of $\omega$) and hence the center of this circle lies on the side $BC$.