ABC is an isosceles triangle, [AD] is an angle bisector
$$\mathrm{m}(\widehat{\mathrm{ACB}}) = 40^{\circ}$$
$$\mathrm{m}(\widehat{\mathrm{ADC}}) = x$$
In the isosceles triangle ABC above where $|\mathrm{AC}| = |\mathrm{BC}|$, what is x in degrees?
A) 105
B) 110
C) 115
D) 120
E) 125