Consider the triangle ACD.
By the theorem on the sum of the angles of a triangle:
180 ° = ∠CAB + ∠ADC + ∠ACD
180 ° = 19 ° + ∠ADC + ∠ACD
∠ADC + ∠ACD = 161 °
Since AD = AC, this triangle is isosceles.
Then, ∠ADC = ∠ACD (by the property of an isosceles triangle), we obtain that:
∠ADC = ∠ACD = 161 ° / 2 = 80.5 °
∠DCB = ∠ACB-∠ACD = 160 ° -80.5 ° = 79.5 °
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