3. As shown in Figure 1, in $\triangle A B C$, $\angle A=60^{\circ}, \angle C=80^{\circ}$, the angle bisector of $\angle C$ intersects the external angle bisector of $\angle A$ at point $D$, and $B D$ is connected. Then the value of $\tan \angle B D C$ is ( ).
(A) 1
(B) $\frac{1}{2}$
(C) $\sqrt{3}$
(D) $\frac{\sqrt{3}}{3}$
3.D.

It is known that $D$ lies on the external angle bisector of $\angle B$.
Given $\angle A B C=40^{\circ}$, we have $\angle D B A=70^{\circ}, \angle D B C=110^{\circ}$.
Also, $\angle B C D=40^{\circ}, \angle B D C=30^{\circ}$, so $\tan \angle B D C=\frac{\sqrt{3}}{3}$.