In a circle centered on the O AC and BD diameters. The angle ACB is 26 ° Find the angle AOD.

Diameters AC and BD are halved by point O. Since AC and BD are diameters, AC = BD. Thus, AC and BD at their intersection form two identical isosceles triangles – △ AOD = △ BOC.

1. Consider △ BOC: ∠OCB (aka ∠ACB) = 26 ° (by condition). Since △ BOC, then ∠OCB = ∠OBC = 26 °.

By the theorem on the sum of the angles of a triangle:

∠OCB + ∠BOC + ∠OBC = 180 °;

26 ° + ∠BOC + 26 ° = 180 °;

∠BOC = 180 ° – 52 °;

∠BOC = 128 °.

2. When the diameters AC and BD intersect, pairs of vertical angles are formed – ∠BOA = ∠COD, ∠BOC = ∠AOD.

Since ∠BOC = 128 °, then ∠AOD = 128 °.

Answer: ∠AOD = 128 °.