Problem

Source:

Tags: geometry, perpendicular bisector



Let $ABC$ be an acuteangled triangle with $AB> BC$ inscribed in a circle. The perpendicular bisector of the side $AC$ cuts arc $AC,$ containing $B,$ in $Q.$ Let $M$ be a point on the segment $AB$ such that $AM = MB + BC.$ Prove that the circumcircle of the triangle $BMC$ cuts $BQ$ in its midpoint.