Let $Z$ be the intersection of $BY$ and $AX$ $\angle XBC = \angle A$ since $CX$ is tangent to $k$. In addition $\angle X = \angle AXT + \angle AXB = \angle ABT + \angle ATB = \angle C$ Therefore $\triangle ABC$ & $\triangle BCX$ are similar $\Rightarrow \angle CBX = \angle B$ Let $M$ be the midpoint of $AC$. It is easy to see that $\angle MBY = \angle B \Rightarrow MBYC$ is cyclic. Therefore $\angle MYB = \angle C$. But $MY // AX $ since $M , Y$ midpoints of $AC , CX$ respectively $\Rightarrow \angle AZB = \angle MYB = \angle C \Rightarrow ABZC$ cyclic. Thus $Z$ lies on $k$ and we are done !

Attachments:

A similar solution as @above works by using of the symmedian $AT$ (i.e. $\angle ABT = \angle CBM$) instead of the cyclic $MBYC$.

Suppose that $AX$ intersects $(ABC)$ at $U$ and $N$ is midpoint of $CA$. We have $\angle{BCX} = \angle{BAC}$ and $\angle{CXB} = 180^{\circ} - \angle{BAT} = \angle{ACB}$. Then $\triangle ABC \cup N \sim \triangle CBX \cup Y$. So $\angle{CBY} = \angle{ABN} = \angle{CBT} = \angle{XBT} - \angle{XBC} = \angle{XAT} - \angle{ABC} = \angle{XAT} - \angle{CAT} = \angle{CAX} = \angle{CBS}$ or $B, S, Y$ are collinear

Suppose that AX intersects (ABC) at Z .We want to prove BZ bisects CX.Firstly <ABT=a,<CBT=b and <CBZ=c.From ABXT is cyclic <ABT=<AXT=a.AT and CT touch (ABC) then <CAT=<ACT=a+b and <ZBC=<ZCX=c.From ABXT is cyclic <TAX=<TBX then <ZBX=a.So we get Z is humpty point in triangle BCX.Then BX is median on triangle BCX.We are done.

Attachments:

Sketch: Angle chase to get $CA$ tangent to $(XBC)$- from here, invert about $C$ at which point the problem becomes vulnerable to complex bash.