I like to complicate questions, so:
Let T{ABC} denote "Triangle ABC" and A{ABC} "Angle ABC".
You can prove T{BAG} is congruent to T{ADE} through SAS.
Construct a line from M through A to point X on BG.
T{AMD} is isosceles (AM = DM, equal radii, part)
hence A{MAD} = A{ADM} (base angles of isos.T{AMD})
Also, A{XBA} = A{ADM} (corresponding angles in congruent triangles)
Therefore A{XBA} = A{MAD}
A{BXA} + A{XBA} = A{MAD} + 90° (exterior angle of T{BAX})
Therefore A{BXA} = 90°