BMO Shortlist, 2019

Let $D$ be the foot of the altitude from $A$ to $BC$, which also lies on $\Gamma$. Let $O$ be the circumcentre of $\triangle ABC$. Since $\angle HDM = 90^\circ$, note that rays $HD$ and $HM$ meet the circumcircle at points which are reflections in $OM$. Then, since $\angle BAD = \angle OAC$, we recover the well-known fact that ray $HM$ meets the circumcircle at $A'$, the point antipodal to $A$. Therefore, the ray $MH$ meets the circumcircle at a point $T$ such that $\angle MTA = 90^\circ$. Note that $T$, $D$ lie on the circle with diameter $AM$.

Figure 4: G4

Now, study $K$, the centre of $\Gamma$. Clearly $AXKY$ is cyclic, with diameter $AK$, so $T$ also lies on this circle. We can now apply the radical axis theorem to the three circles $\odot ATXKY$, $\odot ATDM$, $\odot HXDMY$ to deduce that $AT$, $XY$, $DM$ concur at a point, $Z$.

Then, by power of a point in $\odot ATXY$, we have $ZX \cdot ZY = ZT \cdot ZA$; but also by power of a point in the circumcircle, we have $ZA \cdot ZT = ZB \cdot ZC$. Therefore
$$ZX \cdot ZY = ZB \cdot ZC,$$
and the result follows. $\square$