My solution: Let $ O $ be the circumcenter of $ \triangle ABC $ . Since $ O $ lie on the perpendicular bisector of $ AT $ , so $ O $ also lie on the perpendicular bisector of $ MK $ , hence from symmetry we get $ \triangle OAM $ and $ \triangle OTK $ are congruent . Since $ TK $ is the reflection of $ AM $ in the angle bisector $ AT $ of $ \angle BAC $ , so we get $ TK \parallel AC $ . ... $ (\star ) $ Since $ \angle OKT=\angle OMA=90^{\circ} $ , so combine with $ (\star ) $ we get $ OK \perp AC $ . i.e. $ K $ lie on the perpendicular bisector of $ AC $ and $ AK=KC $ Q.E.D

Let $N$ be on $AC$ such that $CN=AM$.Now,we have that triangles $TMB$ and $TNC$ are congruent so $TM=TN=AK$,so we need to prove $TN=CK$,but since $CN=AM=TK$,it is enough to prove that $CNKT$ is a parallelogram and this is equivalnet showing $TK//AC$,which is an easy angle chase.

Let $S$ be the mid point of side $AC$ and let $O$ be the circumcenter of $ \triangle ABC $ . Clearly $OS$ and $OM$ are perpendicular bisectors of $AC$ and $AB$ respectively. Since $AT\parallel MK$, $\angle BMK$ = $\angle BAT$ = $\angle A/2$ (since clearly $AT$ is the internal bisector of $\angle A$). So, $\angle KMO$ = $90^{\circ} - \angle BMK$ = $90^{\circ} - \angle A/2$ Also since $O$ lies on the perpendicular bisector of $MK$, we have that $\angle MOK$ = $\angle A$. Clearly, $\angle MOA$ = $\angle C$ and $\angle AOS$ = $\angle B$ (from central angle theorem). Hence, $\angle MOK$ + $\angle MOA$ + $\angle AOS$ = $\angle A$ + $\angle C$ + $\angle B$ = $180^{\circ}$. So $KOS$ is a straight line, hence $K$ lies on the perpendicular bisector of $AC$ implying $AK = KC.$

Attachments:

Let's extend line $TK$ to intersect circumcircle of $ABC$ in point $D$. Now we have: $\angle ABD=\angle ATD=\angle BAT=\angle CAT \implies ATBD$ is an isosceles trapezoid $\implies$ $BT=AD, AT \parallel MK \parallel BD$ Now, because $AT \parallel MK \parallel BD$ and because point $M$ is the midpoint of $AB \implies K$ is the midpoint of $DT$. Again we have $\angle CAT=\angle BAT=\angle ATD \implies ACTD$ is also an isosceles trapezoid and since $AD=CT$ and $K$ is a midpoint, it means that $AK=KC$. That is all

Let $R$ be intersection of $(ABC)$ and $KT$. Thus, $\measuredangle ATR=\measuredangle ATK=\measuredangle MAT=\measuredangle BAT\implies ARBT$ is an isosceles trapezoid. Hence, $RK=KT$. Also, $\measuredangle ATR=\measuredangle BAT=\measuredangle TAC\implies TR\parallel AC$, hence $ACTR$ is also an isosceles trapezoid. As $K$ is the midpoint of $ACTR$, $K$ lies on the perpendicular bisector of $AC$, we conclude that $AK=KC$. [asy][asy] size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,M,T,K,R; O=(0,0);A=dir(110);B=dir(200);C=dir(340);path w=circumcircle(A,B,C);T=intersectionpoints(w,A--100*incenter(A,B,C)-99*A)[1];M=midpoint(A--B);R=intersectionpoints(circle(T,abs(A-B)),w)[1];K=midpoint(T--R); draw(A--B--C--cycle,deep);draw(w,deep);draw(A--K--C,org);draw(R--T,deep);draw(M--K,deep);draw(A--T,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$T$",T,dir(T)); dot("$M$",M,dir(M)); dot("$R$",R,dir(R)); dot("$K$",K,dir(K)); [/asy][/asy]

s

Let $O$ be the circumcenter and $N$ be the midpoint of $AC$, then $\angle AON=\angle B$, $\angle MOA=\angle C$. Now we only need $\angle MOK$, we know that $\angle MOK=360^\circ-\angle MOA-\angle KOT-\angle AOT=360^\circ-\angle A-2\angle B-2\angle C=\angle A$. So $\angle KON=180^\circ$, so $K$ lies on perpendicular bisector of $AC$, we're done.