In a circle centered at point 0, diameters AD and BC are drawn, the angle OAB is 70. find the angle value of the OCD.

By the statement of the problem, point O is the center of the circle, AD and BC are diameters, ∠ OAB = 70 °.
Lines OA, OB, OC and OD are the radii of the given circle.
Therefore, Δ AOB and Δ COD are isosceles triangles.
We have: ∠ ABO = ∠ OAB = 70 °.
In a triangle, the sum of the interior angles is 180 °: ∠ AOB + ∠ ABO + ∠ OAB = 180 °, whence ∠ AOB = 40 °.
The vertical angles are: ∠ AOB = ∠ COD = 40 °.
We have: ∠ СOD + ∠ CDO + ∠ OCD = 180 °, whence ∠ OCD = (180 ° – 40 °) / 2 = 70 °.
Answer: 70 °.