A circle with radius $AB$ and center at $A$ is constructed. $D$ is on the circle $CD$ is the angle bisector of $\angle BCA$. $E$ is on the circle so that $DE$ is the angle bisector of $\angle BDA$. Find $\angle BDA$ if $BC \cong CE$.
My work I have done so far is $$a = \angle ADE$$ $$EA \cong AD$$ $$\angle AED \cong \angle ADE$$ $$\angle AED = a$$ $$\angle BDE = a$$ $$\frown BDE = 2a$$ $$\angle EAB = 2a$$ and I'm stuck at this step and I can't continue.
Since $EC = CB$ we have $\angle EDC = \angle CDB = x$, then $\angle ADE = 2x$.
Also $\angle ACD = \angle DCB =y$. Cleraly $$ y = \angle ACD = \angle ADC = 3x$$
Also $$ \angle CAB = 2\angle CDB = 2x$$ so if we look at triangle $ABC$ we have $$4y+2x=180^{\circ} \implies 14 x= 180^{\circ}\implies x = {90^{\circ}\over 7}$$ Finally $$\angle BDA = 4x ={360^{\circ}\over 7}$$