Angle-chasing Olympiad preparation problem, prove A-M-N are collinear

Question

Let ABC be an acute triangle with altitudes AD, BE, CF and circuncenter O. AD and EF intersect at X. AO and BC intersect at Y. Let M and N be the midpoints of XY and BC, respectively. Prove that A-M-N are collinear. I tried some angle chasing and reflecting the Orthocenter H over N to get the antipode of A on the circuncircle, A'. This leaves to prove that AA'H and AYX are similar, or equivalently DXY and DHN are similar, or XY and HN are parallel. Any of these would finish the proof. The main problem is that I don't know how to angle chase to finish the problem. If you find something, thank you.

Answer 1

Thanks to the hint, I was able to complete the proof:

We have that ABA'C is similar to AEHF because BAC=EAF AEH=ABA' AFH=ACA' FHE=CA'B. So, the equivalent of X in AEHF is Y in ABA'C, so AH/AX = AA'/AY and because of that and its positions, AXY~AHA', so the midpoints of the sides XY and HA', M and N are collinear with the common vertex, A. (If you know a way I could improve my proof tell me, please)