Example 2 As shown in Figure 2, in $\triangle A B C$, $A B=A C$, $\angle B A C=100^{\circ}$, point $D$ is on the extension of side $A B$, and $A D=B C$. Find the measure of $\angle B C D$. ${ }^{[2]}$
Solve as shown in Figure 2, construct an equilateral $\triangle A C E$, and connect $D E$.
Then $C E=C A=A E, \angle E A C=60^{\circ}$.
Thus, $\angle D A E=\angle B A C-\angle E A C$
$=40^{\circ}=\angle A B C$.
Also, $A E=A C=A B, A D=B C$, then
$\triangle D A E \cong \triangle C B A$.
Therefore, $D E=A C=C E$,
$\angle D E A=\angle C A B=100^{\circ}$.
And $\angle D E C=\angle D E A+\angle A E C=160^{\circ}$, so $\angle D C E=\angle E D C=10^{\circ}$.
Hence, $\angle B C D=\angle A C E-\angle A C B-\angle D C E=10^{\circ}$.
[Comment] Note that, for an isosceles triangle with a vertex angle of $100^{\circ}$, the base angles are $40^{\circ}$, and their difference is $60^{\circ}$. This is used to construct an equilateral triangle, making " $A D=B C$ " effective through congruent triangles.