Find the angle in the given quadrilateral
The best proof by far:
$B$ is clearly the $D$ excenter of $\triangle ADC$, therefore $DB$ is an angle bissector of that triangle.
So: $\angle CDB = \frac{180^{\circ} - 60^{\circ} - 40^{\circ}}2 = 40^{\circ}$
$\angle DBC = 180^{\circ} - 40^{\circ} - 120^{\circ} = 20^{\circ}$