Taking our point of reference at $T$ and extending $AT,CT$ to intersect $(ABC)$ gives us the following equivalence: Let the altitudes $BE,CF$ of $ABC$ meet at $H$. Let $M,N$ denote the midpoints of arc $FE$ such that $M$ lies on the same side of $EF$ as $A$. Suppose $K$ lies on $MH$ such that $\angle FKE=90^{\circ}$ and $K$ lies on the same side of $EF$ as $A$. Prove that $A,N,K$ are collinear. Proof: Since $AFE\sim ABC$, construct $M'$ in $AFE$ such that it corresponds to $M$ in $ABC$. This implies $AM,AM'$ are isogonal wrt $\angle A$. Since $\angle ECM=\angle FBM$ because $M$ is the midpoint of arc $EF$, and $M,N$ are symmetric about the midpoint of $BC$, therefore it is well known that $AM,AN$ are isogonal. Thus $M'\in AN$. Once we show that $K\equiv M'$, the equivalence follows. Since $\angle FM'E=\angle BMC=90^{\circ}$, it suffices show $M'\in HM$. Let $BM\cap HF=X, CM\cap HE=Y$. Note that $\angle AEM'=\angle ABM=\angle ACM\implies CM||EM'$, and $BM||FM'$ similarly. Since $BX,CY$ are angle bisectors in similar triangles $BHF, CHE$ respectively, therefore \[\frac {FX}{XH}=\frac {YE}{YH}\implies XY||EF\]It follows that $EFM',YXM$ are homothetic about $H$, so $H,M,M'$ are collinear and we are done.

Let bisector $BM$ (external) meet $BC$ at point $N$ and let $NK$ meet $MS$ at point $J$. We claim that $NK \perp MS$. This follows since an inversion about $B$ of radius $\sqrt{BA.BC}$ maps $S$ to $K$ and $M$ to $N$ giving that $MS$ and $NK$ are anti-parallel in angle $ABC$ and since $\angle MBL=90$ we have that $NK \perp MS$. Now, by radical axis theorem, we get that $L$ lies on $(AKC)$. Indeed, $NA.NC=NB.NM=NK.NJ$ and we get the claim. Now, clearly $\angle KJS=90$ implies that $JS$ passes through the anti-pose of $K$ in $AKC$. This proves that $M,T,S,J$ are collinear.

Can anyone post non-inversive solution please? Thanks!!! One thing İ proved is that: The center of $(AKTC) $ is the intersection of $AA $ and $CC $.