58th Ukrainian National Mathematical Olympiad

Let $Z$ be the point of intersection of circle $w$ with line $BC$, then $AZ$ is the diameter of $w$ (see Fig. 35).

Really, if $K=Z$, $w$ is tangent to $BC$, so, as $AK\perp BC$, the center of $w$ is on $AK$. If $K\ne Z$, then $\angle AKZ=90^{\circ }$ and $AZ$ is the diameter of $w$. Then $\angle AMZ=90^{\circ }$.

Let line $MZ$ intersect the circumscribed circle of $△XAH$ a second time at point $S$, then $\angle AMS=90^{\circ }$. Also $\angle SHA=\angle SMA$, then $SH\perp AH$. So $AX\parallel SH\parallel BC$, it also follows that $XAHS$ is a rectangle. So, $AX=SH$.

Also notice that $CH\perp AB$. Apart from that, $ZM\perp AB$. So $HC\parallel SZ$. Then $SHCZ$ is a parallelogram, and so $SH=ZC$. Then $AX=SZ$. Also, $AY=BZ$, so $XY=BC$.