In a circle centered at point O, diameters AD and BC are drawn, and the angle ABO is 80 °. Find the ODC angle

Consider the triangle AOB. This triangle is isosceles, because OA and OB are the radii, so they are equal.
By the property of an isosceles triangle, ∠OAB = ∠ABO = 80 °.
Consider the triangles AOW and COD. ∠DOC = ∠AOB, because they are vertical. СО = DO = OB = OA, because these are the radii of a circle.
Consequently, the triangles AOW and COD are equal (by the first sign). Therefore, ∠OBA = ∠OAB = ∠ODC = ∠OCD = 80 °
Answer: ∠ODC = 80 °