In triangle ABC, the angle A is 55 °. Inside the triangle, point O is marked so that the angle AOB

In triangle ABC, the angle A is 55 °. Inside the triangle, point O is marked so that the angle AOB is equal to the angle of COB and AO = OC. Prove that line BO is the perpendicular to the AC side.

In triangles AOB and COB, the side OB is common, AO = OC, the angle AOB = COB, then the triangles are equal on two sides and the angle between them, then the angle ABO = CBO, which means AB = BC, and then the triangle ABC is isosceles. Then BО is the bisector of triangle ABC, and the segment BH is its height and median, which means AH = CH, BH is perpendicular to AC, and then BО is the midpoint perpendicular to AC, which was required to prove.