Find the measure of B

Let $\angle ABD = x^{\circ}.$ By the Law of Sines,

$\begin{aligned} \dfrac{AD}{BD} &= \dfrac{\sin x^{\circ}}{\sin (150 - x)^{\circ}} \\ \dfrac{AD}{AC} &= \dfrac{\sin 20^{\circ}}{\sin 150^{\circ}}. \end{aligned}$

Since $AC = BD,$ we have

$\begin{aligned} \dfrac{AD}{BD} &= \dfrac{AD}{AC} \\ \dfrac{\sin x^{\circ}}{\sin (150 - x)^{\circ}} &= \dfrac{\sin 20^{\circ}}{\sin 150^{\circ}} \\ \dfrac{\sin x^{\circ}}{\sin (x + 30)^{\circ}} &= 2 \sin 20^{\circ} \\ \dfrac{\sin x^{\circ}}{\sin (x + 30)^{\circ}} &= \dfrac{\sin 40^{\circ}}{\cos 20^{\circ}} \\ \dfrac{\sin x^{\circ}}{\sin (x + 30)^{\circ}} &= \dfrac{\sin 40^{\circ}}{\sin 70^{\circ}} \\ x &= \boxed{40}. \end{aligned}$