Find $x$ angle in triangle

Construct an equilateral triangle such that its sides are equal to the base of the main triangle.

And my second solution is as follow:

Let in $\Delta ABC$ we have $AB=AC$, $\measuredangle A=20^{\circ}$ and $\measuredangle ADC=x$ as on your picture.

Let $M\in AB$ such that $AD=MD$ and $K\in DC$ such that $MK=AD$.

Also, let $B'\in MB$ such that $MB'=AD$ and $C'\in KC$ such that $B'C'||BC.$

Thus, $$\measuredangle MKA=\measuredangle MDK=2\cdot20^{\circ}=40^{\circ}$$ and from here $$\measuredangle B'MK=40^{\circ}+20^{\circ}=60^{\circ},$$ which says $$B'K=MB'=AD=BC.$$ But $$\measuredangle B'KC'=60^{\circ}+20^{\circ}=80^{\circ}=\measuredangle BCA=\measuredangle B'C'A.$$

Thus, $$B'C'=B'K=AD=BC,$$ which says that $$B\equiv B'$$ and $$C\equiv C'.$$ Id est, $$\measuredangle BDC=10^{\circ}+20^{\circ}=30^{\circ}.$$