Диаметры АВ и CD окружности пересекаются в точке О. Найдите величину угла ADO, если LBOD=126°.

The question is a geometry problem involving a circle. Diameters AB and CD intersect at point O. We are given that angle BOD is 126 degrees and asked to find the measure of angle ADO. 1. Identify the relationship between angles: Angles BOD and AOC are vertical angles, so they are equal. Angles BOC and AOD are also vertical angles and are equal. Angles BOD and BOC are supplementary, meaning they add up to 180 degrees. 2. Find angle BOC: Since AB and CD are diameters, they form straight lines. Therefore, angle BOD and angle BOC are adjacent angles on a straight line, making them supplementary. Angle BOC = 180° - Angle BOD Angle BOC = 180° - 126° Angle BOC = 54° 3. Consider triangle ADO: OA, OD, and AD are sides of triangle ADO. OA and OD are radii of the circle, so they are equal in length. Triangle ADO is an isosceles triangle with OA = OD. 4. Find angle AOD: Angles BOC and AOD are vertical angles, so they are equal. Angle AOD = Angle BOC Angle AOD = 54° 5. Find angles OAD and ODA in triangle ADO: In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, angle OAD = angle ODA. The sum of angles in a triangle is 180°. Angle OAD + Angle ODA + Angle AOD = 180° Let angle ADO = x. Then angle OAD = x. x + x + 54° = 180° 2x + 54° = 180° 2x = 180° - 54° 2x = 126° x = 126° / 2 x = 63° Therefore, angle ADO = 63°. Final Answer: The final answer is $$\boxed{63}$$