https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml

Let H' is refection of H using line AB, and point D same as before.
HMc=McD (my Euler line proof 1.) and HC'=C'H' (reflection property) => C'Mc is middle line of triangle H'HD so C'Mc parallel with H'D.
Angle HC'Mc is right => HH'D is also right.
Triangle CH'D is right, CD is diameter => H' is on circumcircle (Thales)

It is not important what type of triangle is ABC.
Easy to prove and vice versa.

THANKS FOR YOUR COMMENT ANOTHER!

https://www.cut-the-knot.org/triangle/ReflectionsOfOrthocenter.shtml

Thank you.