Neat problem! Let $M$ be the midpoint of $AB$. Notice that $X-Y-M$ colinear and parallel to $AD,$ $BC$. Notice that $\angle BAY=\angle XAD=\angle AXY$ and thus $Y$ must be the $X-$Humpty in $\Delta XAB$. This finishes since $YA/AX=YB/BX$.

Wow! Nicely disguised config geo problem in EGMO TST. Let's not post a blatant solution, but rather walk through how you could have approached it in the contest. Just looking at the given angle condition, we could see that $AX$ is the $A-$symmedian of $\triangle ADB$. Its also well known to the contestants that $XY$ is precisely the mid-line of the trapezium $ABCD$. So, we get to know that $XY$ strikes the midpoint of $\overline{AB}$ as well. Suddenly, we start to recall the configuration of the symmedians. Now its often a good strategy to use the angle-bisector theorem(to show that two angle bisectors concur on a line) on $\triangle AYX$ and $\triangle BYX$, it suffices to show $\frac{AY}{AX} = \frac{BY}{BX} \iff \frac{AY}{BY} = \frac{AX}{BX}$. We will now try to show the equality of ratios. Its often a good idea to work with ratios if you have symmedians in your picture. Just let $M$ to be the midpoint of $\overline{AB}$. Trivially, $\angle YAB = \angle XAD = \angle AXY$. So... $MA$ is a tangent to $\odot(XYA)$! Apply Power of a Point with the fact $MA = MB$, to get that $BM$ is tangent $\odot(BXY)$. This would give $\angle YXB = \angle MBY$. But this means $Y$ is the $X-$Humpty point of $\triangle XAB$. But now the ratio which we were trying to chase, is precisely equal by a well-known property of harmonic quadrilaterals and Humpty Point. TST people mostly know the proof, but for pedestrians crossing by, here's a short proof.

Hopefully, you liked the walkthrough!

not dissimilar to above solutions

Good problem