@below You are right. I got confused here.

(In reply to zuss77) T only lies on the reflection of circumcircle about AC. Easy to see, that if T is orthocenter, ABC is equilateral.

Ok, back to this. If $O,H$ - circumcenter and orthocenter of $ABC$ then $AHTOC$ is cyclic. Since $BT$ bisects $\angle ATC$ and $O$ - midpoint of the arc $ATC$ we have $\angle BTO=90^\circ$. Then $T\in (BOEF)$, where $E,F$ - midpoints of $BC,AB$. Note that $M$ is a center of homothety $(ACH)\mapsto (EFO)$, so it maps $T$ and $K$. And since ratio of this homothety is $-1/2$, we have $TM/MK=1/2$.

Another solution: Define $K$ as the point on ray $TM$ such that $\frac{TM}{MK}=\frac{1}{2}$. We will prove that $K$ belongs to the circumcircle of $\triangle{ATC}$ by proving that $\triangle{AKC}$ and $\triangle{BTA}$ are similar. To this end, it suffices to show that: $$\frac{AK}{BT}=\frac{KC}{TA}=\frac{AC}{BA}$$Let $D$ be the intersection of $KN$ and $BT$, where $N$ is the midpoint of $AC$. We have that $\frac{MT}{MK}=\frac{MN}{MB}=\frac{1}{2}$, so $TN\parallel{BK}$. Also, $\frac{TN}{BK}=\frac{TM}{MK}=\frac{1}{2}$. $TN\parallel{BK} \Longleftrightarrow \frac{DT}{DB}=\frac{DN}{DK}=\frac{TN}{BK}=\frac{1}{2}$, so $N$ is the midpoint of $DK$ and $T$ is the midpoint of $BD$. Firstly, we will prove that $\triangle{ATD}$ and $\triangle{DTC}$ are similar to $\triangle{ABC}$. We have that $\angle{BAT}=\angle{BAC}-\angle{TAC}=\angle{BAC}-\angle{TSC}=\angle{BAC}-(180^\circ-\angle{SBC}-\angle{ACB}-\angle{ACS})=\angle{BAC}-(180^\circ-\angle{SBC}-\angle{ACB}-60^\circ)=\angle{BAC}-(180^\circ-\angle{SBC}-\angle{ACB}-\angle{ABC})=\angle{BAC}-(\angle{BAC}-\angle{SBC})=\angle{SBC} \Longleftrightarrow \angle{BAT}=\angle{SBC}$ This, along with the fact that $\angle{ATB}=\angle{BTC}=120^\circ$ gives that $\triangle{ATB}$ and $\triangle{BTC}$ are similar. Hence: $$\frac{AT}{BT}=\frac{AB}{BC}$$$$\frac{TB}{TC}=\frac{AB}{BC}$$Since $T$ is the midpoint of $BD$, we have that: $$\frac{AT}{TD}=\frac{AB}{BC}$$$$\frac{TD}{TC}=\frac{AB}{BC}$$These relationships, along with the fact that $\angle{ATD}=\angle{DTC}=60^\circ=\angle{ABC}$ gives that $\triangle{ATD}$ and $\triangle{DTC}$ are similar to $\triangle{ABC}$. Now, note that $AKCD$ is a parallelogram. Therefore: $$\frac{AK}{BT}=\frac{DC}{DT}=\frac{AC}{BA}$$because $\triangle{DTC}$ is similar to $\triangle{ABC}$. Similarly: $$\frac{KC}{TA}=\frac{AD}{AT}=\frac{AC}{BA}$$because $\triangle{ATD}$ is similar to $\triangle{ABC}$. Thus, we've proven that: $$\frac{AK}{BT}=\frac{KC}{TA}=\frac{AC}{BA}$$which means that $\triangle{AKC}$ and $\triangle{BTA}$ are similar. So, $\angle{AKC}=\angle{ATB}=\angle{ATC}=120^\circ$, which proves that $K$ belongs to the circumcircle of $\triangle{ATC}$. The proof is complete.

Could you put the proper angle symbol here? $\angle$ instead of < Thanks!