6. Convex quadrilateral $A B C D$ satisfies $\angle C A B=\angle A D B=30^{\circ}, \angle A B D=77^{\circ}, B C=C D$, and $\angle B C D=n^{\circ}$ for some positive integer $n$. Compute $n$.
Answer: $\square$
Solution: Let $O$ be the circumcenter of $\triangle A B D$. From $\angle A D B=30^{\circ}$, we get that $\triangle A O B$ is equilateral. Moreover, since $\angle B A C=30^{\circ}$, we have that $A C$ bisects $\angle B A O$, and thus must be the perpendicular bisector of $B O$. Therefore, we have $C B=C D=C O$, so $C$ is actually the circumcenter of $\triangle B D O$. Hence,
$$
\begin{aligned}
\angle B C D & =2\left(180^{\circ}-\angle B O D\right) \\
& =2\left(180^{\circ}-2 \angle B A D\right) \\
& =2\left(180^{\circ}-146^{\circ}\right)=68^{\circ}
\end{aligned}
$$