Let the line $AQ$ intersect the circumcircle of $BOC$ at point $Q$ and $T'$. Notice that $\overline{OQ}$ is a diameter of this circle, so $\angle OT'Q = 90^\circ$.
Let $P$ be the midpoint of the segment $\overline{AC}$. Since $O$ is the centre of the circumcircle of $ABC$, we have $\angle OPA = 90^\circ$.
From here, we get $\angle OT'A = 180^\circ - \angle OT'Q = 90^\circ = \angle OPA$, which implies that the quadrilateral $AOT'P$ is cyclic.

Notice that point $P$ lies on the line $MN$ (moreover, $P$ is the midpoint of $\overline{MN}$) because $ANCM$ is a parallelogram. Thus,
$$\angle AT'M = \angle AT'P = \angle AOP = \frac{1}{2} \angle AOC = \angle ABC.$$
As in the previous solution, we show that the quadrilateral $ABNT'$ is cyclic and that the points $M, N$ and $T'$ are collinear. This implies $T \equiv T'$, which concludes the proof.