Pr. 8537 Romantics of Geometry, Konstantin Knopp, https://www.facebook.com/photo?fbid=10222355364046501&set=gm.4147069368740071

 Let G be a point inside the cyclic hexagon ABCDEF so that <AGB=60, <EGC=90. If H is midpoint of BC, then G lies onto FH.

Proof:
If O is the center of the hexagon, G is intersection of the circles (AOB) and (CE). Noticing that AF is tangent to the circle (ABO), EF tangent to the circle (CE); as AF=EF, F belongs to the radical axis of the 2 circles. Noticing that BC is common tangent to the 2 circles, H belongs to their radical axis as well, and we are done. There are 2 internal points G, G', determining the radical axis FH of the 2 circles.