Let $T$ be the midpoint of $XY,\ S=AB\cap XY$, and $P=d\cap XY$. If we are to prove that $XY$ passes through a fixed point, then that point lies on the limit of $XY$ as the radius of $\omega$ goes to $0$, and that line is the perpendicular $\ell$ through $B$ to $AB$. Proving that $XY\cap\ell$ is constant is equivalent, because $AT\perp XY$, to proving that $T$ moves on a circle arc passing through $A,B$ (and then the desired fixed point will be the antipode of $A$ in this circle). This means that it suffices to prove that $\angle ATB$ is constant, or, equivalently, that $\angle PTB$ is constant. $BS$ is the $B$-symmedian of $BXY$, so $\angle YBS=\angle TBX$. This means that $\angle PBS=\angle PBY+\angle YBS$, which equals $\angle BXT+\angle TBX=\angle PTB$. This shows that $\angle PTB=\angle PBS=\angle PBA$, which is, of course, constant.

during the contest i managed to solve it using coordinates. the computations are very easy... i took $B$ as origin and $d$ as y-axis, and the equation of $XY$ was a first degree expresion, with one parameter. after i managed to remember the equation of tangent to circle, the problem took me 5 minutes...

Let $Q$ be the intersection of the perpendicular at $B$ to $AB$ with $XY$. It's interesting to notice that the midpoint of $AQ$ lies on $d$. This follows from the fact that $\triangle{OTB}\sim\triangle{OBA}$ and hence $\angle{AQB}=\angle{BTO}=\angle{OBA=\angle{PBQ}}$.

Let $ M $ , $ N $ be the midpoints of $ AX $ , $ AY $ respectively.Analogically $ M' $ , $ N' $ for $ X'Y' $ then $ MN $ , $ M'N' $ , $ d $ intersect at one point ( radical axis for A,w,w') now let the homothety at center A and coefficient 2.then $ MN $ goes to $ XY $ , $ M'N' $ goes to $ X'Y' $ and $ d $ goes to $ d' $ because of $ d' $ is constant line and we have that all such $ XY $ intersect at the $ d' $ then we have that all $ XY $ goes through fixed point