6. In $\triangle A B C$, $A B=A C, \angle A=100^{\circ}, I$ is the incenter, $D$ is a point on $A B$ such that $B D=B I$. Find the measure of $\angle B C D$.
《Mathematics Bulletin》Problem 1073
6. Connect $D I, I C$, let $\angle D C I=x$, then $\angle I D C=40^{\circ}-x$, apply the trigonometric form of Ceva's theorem to $\triangle B C D$ and point $I$, we have $\frac{\sin 80^{\circ}}{\sin \left(40^{\circ}-x\right)} \cdot \frac{\sin 20^{\circ}}{\sin 22^{\circ}} \cdot \frac{\sin x}{\sin 20^{\circ}}=1$, solving for $x=10^{\circ}$, hence $\angle B C D=30^{\circ}$ is the answer.