The Ex-black Problem

We note that $\angle ABC =60^\circ$ and $EB=BC$ , therefore $\triangle BCE$ is equilateral, and its three sides and internal angles are equal. We also note that $\angle DBC=45^\circ$ , since $\angle C = 90^\circ$ , $\angle BDC = 45^\circ$ , and $\triangle BCD$ is isosceles with $CD=BC$ . Therefore $\triangle CDE$ is also isosceles, since $CD=EC$ and $\angle CDE = \frac {180^\circ - \angle DCE}2 = \frac {180^\circ - 30^\circ}2 = 75^\circ$ . And $\angle EDB = \angle CDE-\angle BDC = 75^\circ - 45^\circ = \boxed{30}^\circ$ .

Let $|\overline {BC}|=|\overline {BE}|=a$ . Since $\angle {CAB}=30\degree$ , therefore $\angle {ABC}=60\degree\implies \angle {CBD}=45\degree\implies |\overline {BD}|=a\sqrt 2$ . Applying sine rule, $\dfrac{a\sqrt 2}{\sin (15\degree+x)}=\dfrac{a}{\sin x}\implies \tan x=\dfrac{1}{\sqrt 3}\implies x=\boxed {30\degree}$ .