Geometry Problem on $\triangle ABC$ and Angle Chasing

$\angle ABC=\angle ABD+\angle DBC=80^\circ$.

\begin{align*} AB&=BC\\ \implies \angle CAB&=\angle BCA=(180^\circ-\angle ABC)/2=50^\circ. \end{align*}

Erect an equilateral triangle $ACE$ on base $AC$. Then $\triangle$s $ABE, CBE$ are congruent in opposite sense because $AB=CB$, $AE=CE$ and $BE$ is common. Thus $$\angle AEB=\angle BEC=30^\circ.$$

$$\angle CDB=180^\circ-\angle DBC-\angle BCD=150^\circ.$$ Thus quadrilateral $BDCE$ is cyclic because its angles $D$ and $E$ are supplementary. Thus $$\angle DEC=\angle DBC=20^\circ.$$

\begin{align*} \angle ECB&=\angle ECA-\angle BCA=10^\circ\\ \implies \angle ECD&=\angle ECB+\angle BCD=20^\circ=\angle DEC. \end{align*}

Thus triangle $CED$ is isosceles on base $CE$, so $CD=DE$. Thus $\triangle$s $ACD, AED$ are congruent in opposite sense because $AC=AE$, $CD=ED$ and $AD$ is common. Thus

\begin{align*} \angle CAD&=\angle DAE=30^\circ\\ \angle BAE&=\angle CAE-\angle CAB=10^\circ\\ \implies \angle DAB&=\angle DAE-\angle BAE=20^\circ\\ \implies \angle BDA&=180^\circ-\angle DAB-\angle ABD=100^\circ. \end{align*}

Continue to simplify

$$\begin{align} \cot x & =\frac{4\sin 10-1}{\sqrt{3}} =\frac{(2\sin 10-\frac12)\cos10}{\frac{\sqrt{3}}2\cos10} \\ & =\frac{\sin 20-\cos60\cos10}{\cos10\sin60} =\frac{2\cos 70-2\cos60\cos10}{\cot10\cdot2\sin10\sin60} \\ & =\frac{\cos70-\cos50}{\cot10\cdot(\cos50-\cos70)} =-\cot80=\cot100 \end{align}$$

Thus, $x=100^\circ$.

Let $E$ be the circumcenter of $BCD$. Then $\angle BED=2\angle BCD=20^\circ$ and $\angle DEC =2\angle DBC =40^\circ$. Hence $\angle BEC=60^\circ$. This and $BE=EC$ shows that $BEC$ is equilateral. So $BC=BE$ and $\angle CBE=60^\circ$. By assumption $AB=BC$, so $AB=BE$ and $$\angle BEA = 90^\circ -\frac 12 \angle ABE =90^\circ -\frac 12 \cdot 140^\circ =20^\circ =\angle BED.$$ Therefore $A,D,E$ are collinear and we find $$\angle BDA =180^\circ -\angle EDB = \angle BED+\angle DBE= 20^\circ+80^\circ =100^\circ.$$