Isoscleed!

Consider the diagram.

$\angle BDA=180-20-60-50=50$

$\angle AEB=180-60-30-50=40$

Since $\angle ADB=\angle DBA=50$ , $\triangle ADB$ is isosceles with $AD=AB$ .

Let $AD=AB=1$ .

Apply sine law on $\triangle AEB$ .

$\dfrac{AE}{\sin 80}=\dfrac{1}{\sin 40}$ $\implies$ $AE \approx 1.532$

Appy cosine law on $\triangle ADE$ .

$(DE)^2=1^2+1.532^2-2(1)(1.532)(\cos 20)$ $\implies$ $DE \approx 0.684$

Apply sine law on $\triangle ADE$ .

$\dfrac{\sin x}{1}=\dfrac{\sin 20}{0.684}$

$x=\sin^{-1}0.5=$ $\boxed{30^\circ}$

Note:

I did not used the original figure but I used the correct angles.

Observe that angle(BEC)=50°, angle(BDC)=40° and BE=BC. Take the point "F" on DC, such that angle(CBF)=20°. Observe that the triangle BFC is isosceles with BC=BF. Then BF=BE with angle(EBF)=60°, and the triangle BEF is equilateral. As the angle(BDC)=40°, then the triangle BFD is isosceles, so BF=FD, and since EF=BF=BE then DF=EF. The angle(EFD)=40° since angle(BFC)=80°, and the triangle EFD is isosceles; so angle(EDF)=70° and we will angle(EDB)= 30°.