find angle BCD

Let $|\overline {AB}|=|\overline {AC}|=2x, |\overline {BC}|=2y, \angle {BCD}=α$ . Then, applying sine rule to $\triangle {BDC}$ , we get

$\dfrac{y}{\cos B}=\dfrac{2y}{\cos (B-α)}\implies y\cos (B-α)=2y\cos B$ . So,

$|\overline {BD}|=x, |\overline {CD}|=y$ . We have $\angle {ABC}=\angle {ACB}=B, \angle {BAC}=180\degree-2B$ .

Then we have $y\cos B=x(1+\cos 2B)\implies y=2x\cos B$ (from $|\overline {AC}|=|\overline {AE}|+|\overline {EC}|\implies 2x=2x\cos (180\degree-2B)+y\cos (B-α)=-2x\cos 2B+2y\cos B$ ).

Applying sine rule to $\triangle {BDC}$ we get $\dfrac {x}{\sin \angle {BCD}}=\dfrac{y}{\cos B}=\dfrac{2x\cos B}{\cos B}\implies \sin \angle {BCD}=\dfrac{1}{2}\implies \angle {BCD}=\boxed {30\degree}$ .