3. In isosceles $\triangle A B C$, vertex angle $\angle B A C=$ $100^{\circ}$, extend $A B$ to $D$, such that $A D=B C$. Then $\angle B C D$ $=(\quad)$.
(A) $10^{\circ}$
(B) $15^{\circ}$
(C) $20^{\circ}$
(D) $30^{\circ}$
3. (A).

As shown in Figure 2, draw $DE // BC$, intersecting at $DE = AB$, and connect $AE$. It is easy to prove that $\triangle ABC \cong \triangle EAD$.
Thus, $\angle 2 = 40^{\circ}$,
$$
\angle 3 = 60^{\circ}, AE = EC.
$$

Therefore, $\triangle AEC$ is an equilateral triangle. From $\angle 6 = 40^{\circ}$, we get $\angle BCE = 20^{\circ}$. Also, $\angle ADE = 40^{\circ}$, so $DE = AE = EC$. Therefore, $\angle 4 = \angle 5 = \angle BCD = 10^{\circ}$.