Sunaina thinks absurd

  This problem is the same level as last year's P2 or a bit harder, I feel.  No hand diagram because I didn't use any diagram~ (I head solved it) Problem: Let $ABC$ be a triangle with $AC>AB$ , and denote its circumcircle by $\Omega$ and incentre by $I$. Let its incircle meet sides $BC,CA,AB$ at $D,E,F$ respectively. Let $X$ and $Y$ be two points on minor arcs $\widehat{DF}$ and $\widehat{DE}$ of the incircle, respectively, such that $\angle BXD = \angle DYC$. Let line $XY$ meet line $BC$ at $K$. Let $T$ be the point on $\Omega$ such that $KT$ is tangent to $\Omega$ and $T$ is on the same side of line $BC$ as $A$. Prove that lines $TD$ and $AI$ meet on $\Omega$. We begin with the following claim! Points $B,X,Y,C$-are concyclic Because $CD$-is tangent to the incircle, we get that $\angle CYD=\angle BXD$ and $\angle CDY=\angle DXY$. So $$\angle BXD+\angle DXY+YCB=180 \implies \angle BXY+\angle YCB=180.$$  Also note that $K-B-C$ is radical axis of the incircle and $B