In isosceles triangle ABC, the bisector CD of angle C is equal to the base BC. Then the angle CDA is.

Let us denote the angle at the vertex A through a, then since the triangle is an isosceles angle at the base ACB = (180 – a) / 2. Since CD is a bisector, the angle ACD = 1/2 * ACB = (180 – a) / 4. Consider the triangle ACD:

a + (180 – a) / 4 + CDA = 180;

4a + 180 – a + 4 * CDA = 720;

4 * CDA = 540 – 3a;

CDA = (540 – 3a) / 4.