Let $O$ be the center of $\omega$, thus $O=M Y \cap N X$. Let $\ell$ be the perpendicular bisector of $A T$ (it also passes through $O$). Denote by $r$ the operation of reflection about $\ell$. Since $A T$ is the angle bisector of $\angle B A C$, the line $r(A B)$ is parallel to $A C$. Since $O M \perp A B$ and $O N \perp A C$, this means that the line $r(O M)$ is parallel to the line $O N$ and passes through $O$, so $r(O M)=O N$. Finally, the circumcircle $\gamma$ of the triangle $A M T$ is symmetric about $\ell$, so $r(\gamma)=\gamma$. Thus the point $M$ maps to the common point of $O N$ with the arc $A M T$ of $\gamma$ - that is, $r(M)=X$. Similarly, $r(N)=Y$. Thus, we get $r(M N)=X Y$, and the common point $K$ of $M N$ and $X Y$ lies on $\ell$. This means exactly that $K A=K T$. !