In triangle ABC, angle ACB = 55 degrees, BO-median, BO = CO. Find the angle ABO.

Since, by condition, BO = CO, then the BOC triangle is isosceles, which means the angle OBC = ABC = 55.

Then the angle BОС = 180 – ОВС – АСB = 180 – 55 – 55 = 70.

The angles AOB and BOC are adjacent, then their sum is 180, which means that the angle AOB = 180 – BOS = 180 – 70 = 110.

Since BО is the median of the triangle, then OA = OC, and since BO = OC, then OA = OB, and triangle AOB is isosceles, which means the angle BAO = ABO.

Angle BAO = ABO = (180 – AOB) / 2 = (180 – 110) / 2 = 70/2 = 35.

Answer: Angle ABO is 35.