<Tri-2025-2202>, hint

Bisector of <ACB meet AM at N, since Triangles BCM and CNM are similar

MC/MN=BM/MC  MC^2=MN x MB. hence <BCA=90

Triangle BCM is right triangle MC=BD=BC=CD Hence C is the center of circle  

Triangle BCD is equilateral. Since <BCD=60, <BMD=30