In a circle centered at point O, the radii OB and OA are drawn so that ﮮ AOB = 60 °, OB = DС. Find the value ﮮ ADO.

Let’s construct the OС radius of the circle.
Since, by condition, ОВ = DС, then ОВ = ОВ = СD = R, and therefore, the triangle OCD is isosceles, then the angle COD = COD.
The central angle AOB, by condition, is equal to 600, then the arc AB is also equal to 600.
Let the sought angle ADO be equal to X0, then the angle COD = ADO = COH = X0.
The central angle of AOH rests on the CH arc, then the CH = X0 arc.
The segments DA and DB are secant, behaved from one point, then the angle of ADB is equal to the half-difference of the degree measures of the arcs AB and CH.
Angle ADB = ADO = (AB – CH) / 2.
X = (60 – X) / 2.
2 * X = 60 – X.
3 * X = 60.
X = 60/3 = 200.
Answer: The value of the ADO angle is 200.

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