Find the angle in the given quadrilateral 
I've tried to solve it by assuming the angle to be x and I've tried using the following properties:
1) sum of angles of triangles is 180.
2) sum of angles of quadrilateral is 360.
3) sum of supplementary angles is 180.
However, I still haven't been able to figure out the value of x.
 A: The best proof by far:
$B$ is clearly the $D$ excenter of $\triangle ADC$, therefore $DB$ is an angle bissector of that triangle.
So: $\angle CDB = \frac{180^{\circ} - 60^{\circ} - 40^{\circ}}2 = 40^{\circ}$
$\angle DBC = 180^{\circ} - 40^{\circ} - 120^{\circ} = 20^{\circ}$
A: For convenience, all the angles below are in degree.
In $\triangle ACD$, the sine law gives us
$$\frac{CD}{AC} = \frac{\sin 40 }{\sin 80} = \frac 1{2\cos 40}.$$
Similarly, in triangle $ACB$,
$$\frac{BC}{AC} = \frac{\sin 70}{\sin 50} = \frac{\cos 20}{\cos 40}.$$
Thus
$$\frac{CD}{BC} = \frac{1}{2\cos 20} = \frac{\sin 20}{\sin 40}.$$
Now, if we denote by $x$ the angle $\angle DBC$, then $\angle BDC = 60 - x$. Also, the sine law in $\triangle BCD$ yields
$$\frac{BC}{CD} = \frac{\sin x}{\sin (60-x)}.$$
So we have
$$\frac{\sin x}{\sin (60 -x)} = \frac{\sin 20}{\sin 40}.\tag{1}$$
Note that the function
$$f(x) = \frac{\sin x}{\sin (60-x)}$$
is increasing for $0< x < 60$, so (1) gives us $x =20$.
A: Pure euclidean geometry solution:

Construct equilateral triangle ADP as shown in the figure above.
The quadrilateral ADCP is cyclic.
Then you can get the values of $\angle PAB=50, \angle APC=100$ by simple angle chasing
Observe $\triangle APB$ you can find that $\angle ABP = 50$. Then the triangle ABP is isosceles (AP=BP)
As ADP is an equilateral triangle DP=PB
Now let the $\angle PBD=x$ then,
$2x=40, x=20$
