A geometry problem by Aly Ahmed

$\dfrac{|\overline {DC}|}{|\overline {AD}|}=\dfrac{\sin 20\degree}{\sin (110\degree-x)},\dfrac{|\overline {DB}|}{|\overline {AD}|}=\dfrac{\sin 10\degree}{\sin x}. |\overline {DC}|=|\overline {DB}|\implies \sin 20\degree\sin x=\sin 10\degree\sin (110\degree-x)\implies x=\tan^{-1}\left (\dfrac{\sin 10\degree\sin 110\degree}{\sin 20\degree+\sin 10\degree\cos 110\degree}\right ) =\boxed {30\degree}$ .

Take a point $E$ on $AD$ such that $ED =DB= DC$ , thus $E, B , C$ belong to a circle of center $D$ .

$\triangle EDB$ is an isosceles triangle, thus the congruent angles each measures $\angle DBE= \angle DEB = 20^\circ$ .

Also, we have $\triangle ACD , \triangle ADB$ are congruent triangles by S.A.S.

$\triangle CDE, \triangle EDB$ are congruent triangles by S.A.S.

Thus, $\triangle AEC, \triangle AEB$ are congruent by deduction.

Therefore: $EA = EC = EB$ corresponding sides of congruent triangles.

$\angle AEB= 160^\circ \implies \angle EAB= \angle EBA = 10^\circ$

Thus, $x = 20^\circ + 10^\circ = \boxed {30^\circ}$

Also, I forget to mention that the three blue triangles are congruent by S.S.S, which makes the interior angles of this triangle $\angle EDC = \angle EDB = \angle CDB = 140^\circ$