Solution:

Let $O$ be the circumcenter of $\triangle ABD$. From $\angle ADB = 30^{\circ}$, we get that $\triangle AOB$ is equilateral. Moreover, since $\angle BAC = 30^{\circ}$, we have that $AC$ bisects $\angle BAO$, and thus must be the perpendicular bisector of $BO$. Therefore, we have $CB = CD = CO$, so $C$ is actually the circumcenter of $\triangle BDO$. Hence,
$$\begin{aligned} \angle BCD & = 2\left(180^{\circ} - \angle BOD\right) \\ & = 2\left(180^{\circ} - 2 \angle BAD\right) \\ & = 2\left(180^{\circ} - 146^{\circ}\right) = 68^{\circ} \end{aligned}$$