9. As shown in Figure 1, in $\triangle A B C$, it is known that $D$ is a point on side $A C$, $B D=D C, \angle B C D=$ $70^{\circ}$. Then $\angle A D B=$ ( ).
(A) $100^{\circ}$
(B) $120^{\circ}$
(C) $135^{\circ}$
(D) $140^{\circ}$
(E) $150^{\circ}$
9. D.

From $B D=D C$, we know $\angle B C D=\angle C B D$.
Also, $\angle B C D=70^{\circ}$, so $\angle C B D=70^{\circ}$.
Therefore, $\angle A D B=70^{\circ}+70^{\circ}=140^{\circ}$.