Olympiad Geometry Problem #61: Angle Bisector, Perpendicular Bisector Cyclic Quad

Here is a delightful problem from the Saint Petersburg Math Olympiad this very year in 2020! My inspiration for the solution came from another video on my channel. Enjoy! Link below.

[mikesyd]

Amazing solution! I will offer my:
So we draw circle (ADE) and suppose it intersects BC at point V. I will prove that V=F. If we project through point M on your diagram this quadrilateral on (AIC), we will get a symmetrical quadrialteral. But all symmetrical quadrilaterals are harmonic. So ADVE is also harmonic. EDV=EAV=MAD. To prove that V=F, we have to prove that B1DM=EDV <=> B1DM=MAD. SO we have to prove that MD^2=MI*MA, which is obvious

[chetdivedau]

Very nice solution!
Here is other way to show that B, D, M, E are concyclic. No hard to show that the triangleS MAB and MB1A are similar, hence MH.MB=MB1.MB=MA^2=MD.ME so B, D, M, E are concyclic.