ABC is an isosceles triangle with base AC, CD is the bisector of angle C, angle ADC + 150 degrees. Find angle B

1. Half of the angle C is denoted by x. Then C = 2 * x; A = C = 2 * x – as angles at the base.

2. Now let us recall the property of the external angle of a triangle: it is equal to two internal angles, not adjacent to this external one.
Let’s compose a system:
{B + x = 150 °;
{B + 4 * x = 180 °.

Let us subtract the first from the second equality: 4 * x – x = 180 ° – 150 °; 3 * x = 30 °; x = 10 °.
B = 150 ° – x = 140 °.
Check: C = 2 * x = 20 °. A = C = 20 °. B = 180 ° – (A + B) = 140 °,