Replicate to Solve

Let $\angle {ABD}=α$ . Then

$\dfrac{|\overline {BD}|}{\sin 20\degree}=\dfrac{|\overline {AD}|}{\sin α}$

$\dfrac{|\overline {BD}|}{\sin 80\degree}=\dfrac{|\overline {BC}|}{\sin (20\degree+α)}=\dfrac{|\overline {AD}|}{\sin (20\degree+α)}$

$\implies \sin 80\degree\sin α=\sin 20\degree\sin (20\degree+α)$

$\implies \tan α=\dfrac{\sin^2 20\degree}{\sin 80\degree-\sin 20\degree\cos 20\degree}=\tan 10\degree$

$\implies α=\boxed {10\degree}$ .

The solution is very simple and becomes apparent after one notices the construction that can be made.

Construct a triangle EDA on base AD which is congruent to triangle ABC.

Then join point E with B.

Immediately

Angle AED = Angle CAB = 20 degrees

Also Angle EAB = 80 degrees-20 degrees=60 degrees

Observe AE = AB by congruency criterion

and Angle EAB = 60 degrees

Thus triangle EAB is equilateral

Thus BE = AB

Also by congruency criterion applied to triangle ABC and triangle ADE, ED=AB

Thus ED=BE

Also by angle-chasing in triangle EDA and triangle EDB,

Angle BED=40 degrees

since ED=BE, angle EBD=70 degrees(by isosceles triangle theorem and angle sum property in triangle EBD

Angle EBA=60 degrees(equilateral triangle property)

so Angle ABD=70 degrees - 60 degrees=10 degrees

All angles are in degrees.