In triangle ABC, angle C is 2 times the angle B, CD is the bisector. From the middle M of the side BC

In triangle ABC, angle C is 2 times the angle B, CD is the bisector. From the middle M of the side BC, the perpendicular MH is lowered onto the segment CD. On the side AB, there was such a point K that KMH is an equilateral triangle. Prove that points M, H and A are collinear.

Consider triangle ABC: 1. Connect points D and M. We get that DM is the height, median and bisector of triangle DBC, since this triangle is isosceles (angle DCB = angle DMH = 30 °) => AD = DM => DH is perpendicular to AM … 2. Since MA coincides with MN, since from one point (M) only one perpendicular can be drawn to one straight line (DC). So points A, H and M lie on one straight line. Q.E.D.