The correct option is C 15∘
In △ABC
AB=AC
∠B=∠C (isosceles triangle property) (1)
∠ADC=∠B+30 (Exterior angle property) (2)
Also, ∠AED=∠EDC+∠C (Exterior angle property)
∠AED=∠EDC+∠B (From 1)
Now, given, AD=AE
hence, ∠ADE=∠AED (Isosceles triangle property)
Thus, ∠ADE=∠EDC+∠B
∠ADE+∠EDC=2∠EDC+∠B
∠ADC=2∠+∠B
∠B+30=2∠EDC+∠B (From 2)
∠EDC=15∘