TITLE: Find $\angle BDA$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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 $$\angle BDA = 4x ={360^{\circ}\over 7}$$