19. 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 $\angle B C D$.
(Problem 1073 from Mathematical Bulletin)
19. Connect $D I, I C$, from $A B=A C$ and $I$ being the incenter, we know $\angle B C I=\angle I B C=\angle I B D=20^{\circ}$. Also, from $B D=B I$, we know $\angle B D I=80^{\circ}$. Let $\angle D C I=x$, then $\angle I D C=40^{\circ}-x$. For $\triangle B C D$ and point $I$, we have $\frac{\sin 80^{\circ}}{\sin \left(40^{\circ}-x\right)} \cdot \frac{\sin 20^{\circ}}{\sin 20^{\circ}} \cdot \frac{\sin x}{\sin 20^{\circ}}=1$, solving for $x=10^{\circ}$. Therefore, $\angle B C D=30^{\circ}$ is the answer.