Point D on side AB of triangle ABC is chosen so that AD = AC. It is known that ∠CAB = 80 ° and ∠ACB = 59 °. Find the angle of the DCB.

Consider the triangle ACD.
By the theorem on the sum of the angles of a triangle:
180 ° = ∠CAB + ∠ADC + ∠ACD
180 ° = 80 ° + ∠ADC + ∠ACD
∠ADC + ∠ACD = 100 °
Since AD = AC, this triangle is isosceles.
Then, ∠ADC = ∠ACD (by the property of an isosceles triangle), we obtain that:
∠ADC = ∠ACD = 100 ° / 2 = 50 °
∠DCB = ∠ACB-∠ACD = 59 ° -50 ° = 9 °
Answer: 9