13. 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 $\angle B C D$.

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Translation:

13. 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 $\angle B C D$.
13. As shown in the figure, connect $C I$, and construct an equilateral $\triangle E C I$ on the inside of $\triangle I B C$ with $C I$ as one side. Connect $B E$. It is easy to see that $\angle E C B=\angle 40^{\circ}=\angle A B C$, thus $C E \parallel B D$.

Also, $C E=C I=B I=B D$, which means quadrilateral $D B E C$ is a parallelogram, so $\angle B C D=\angle C B E$.

In $\triangle I B E$, it is easy to see that $I B=I E$, and $\angle B I E=80^{\circ}$. Alternatively, from $\angle A + \angle A D I + \angle A C I = 220^{\circ}$, we know $\angle D I C = 140^{\circ}$, thus $\angle B I E = 360^{\circ} - (140^{\circ} - 60^{\circ} - 80^{\circ}) = 80^{\circ}$, which means $\angle I B E = 50^{\circ}$. Therefore, $\angle D C E = 70^{\circ}$, so $\angle D C I = 10^{\circ}$, and thus $\angle B C D = 30^{\circ}$.