Since in any isosceles triangle the angles at the base are equal, and the base of this isosceles triangle is the segment AC, then ∠BAC = ∠BCA.
According to the condition of the problem, the BCA angle is 50 °, therefore, the BCA angle is also 50 °
Consider triangle ADC.
Since in any triangle the sum of the angles is always 180 °, the following relationship takes place:
180 ° = ∠ADC + ∠BAC + ∠BCA = ∠ADC + 50 + 50 = ∠ADC + 100,
whence it follows that ∠ADC = 180 – 100 = 80 °.
Answer: ∠ADC = 80 °.
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