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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.

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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 $x =20$.

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