Let $D'$ be a point on the circle with center $A$ and radii $AB=AC$. and $\angle CAD'=30^{\circ}$. Then $\angle D'BC=\frac{\angle CAD}{2}=15^{\circ}$. Also $\triangle AD'C $ is an isosceles in $A$ thus $\angle ACD=\frac{180^{\circ}-\angle CAD}{2}=75^{\circ}$ whence $\angle BCD'=120^{\circ}$ hence $D'=D$ and so $AB=AD$ q.e.d.