Example 4 As shown in Figure 4, in $\triangle A B C$, $A B=A C, \angle A$ $=100^{\circ}, I$ is the incenter, and $D$ is a point on side $A B$ such that $B D=B I$. Find the measure of $\angle B C D$.

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Solve as shown in Figure 4, connect $D I$ and $I C$.
Given $A B=A C$, and $I$ is the incenter, we know $\angle B C I=\angle I B C=\angle I B D=20^{\circ}$. Since $B D=B I$, we have $\angle B D I=80^{\circ}$. Let $\angle D C I=x$. Then $\angle I D C=40^{\circ}-x$. Applying the trigonometric form of Ceva's theorem to $\triangle B C D$ and point $I$, we get
$$
\begin{array}{l}
\frac{\sin 80^{\circ}}{\sin \left(40^{\circ}-x\right)} \cdot \frac{\sin 20^{\circ}}{\sin 20^{\circ}} \cdot \frac{\sin x}{\sin 20^{\circ}} \\
=1\left(0^{\circ}0, g^{\prime}(x)<0 \text {. }
$$

Thus, if the equation has a solution, the solution is unique.
When $x=10^{\circ}$, the equation holds, therefore, $x=10^{\circ}$.
Hence, $\angle B C D=\angle B C I+\angle I C D=30^{\circ}$.