which side is isos

Note that since $BC = AC$, we have: $$\frac{BC}{\sin \angle BDC} = \frac{AC}{\sin \angle ADC}$$However this implies that (by the sine rule): $$\frac{AD}{\sin \angle ACD} = 2 \cdot AD$$which further implies that $\angle ACD = 30^\circ$.

Let $O$ be the center of $(BCD)$ and let $(BCD)$ intersect $AC$ at $T\neq C$. Then by angle chasing $AD=DT$ and $DO=OT$ so triangle $DTO$ is equilateral. Now $\angle ACD=\angle TCD=\frac{1}{2}\angle TOD =30^{\circ}$.