We claim that whenever points $X$ and $Y$ satisfy the conditions, the circumcircles of $\triangle ABC$ and $\triangle AXY$ are tangent to each other at $A$. Indeed, this is equivalent to $\angle XYA - \angle CBA = \angle XAC$. But we have because of the conditions $$\angle XYA - \angle CBA = \angle CYA - \angle CYX - \angle CBA = \angle AXB - \angle ACB - \angle CBX = \angle XAC.$$ Now, if $D$ denotes the point of intersection of the tangent to the circumcircle of $\triangle ABC$ with $BC$ (it exists because of $AB > AC$ and is independent of $X,Y$), we immediately get by the radical axis theorem that $XY$ passes through $D$.