First of al, by the sum of the angles in ABC,

a+b+c=180

Moreover, by the sum of the angles in AMB,

a/2+b/2+d=180

With a little computation

d=c/2+90

Since c is constant, d is constant. By the chord theorem, M is on a circle section.

By the way the midpoint M* of this circle is also on the circumscribed circle of ABC.

Can you prove this?

End of the tutorial!