Clearly, there is only one such point $E$. Let the circle with center $B$ and radius $BA=BC$ and the circle with center $C$ and radius $CB=CD$ meet at $E'$. We claim that $E' \equiv E$. Notice that $BE'C$ is equilateral. Thus, $\angle BDE' = \frac{1}{2} \angle BCE = 30^\circ$ and $\angle E'AC = \frac{1}{2} E'BC = 30^\circ$ which, by the uniqueness of $E$, means that $E \equiv E'$. So, $\angle BEC = \angle BE'C = \boxed{60^\circ}$.

STAYPOPSIN

Let $\omega: \text{the circle with center B and radius BA}$ and $\Omega: \text{the circle with center C and radius CB}$ Now $E':\omega \cap \Omega$ But, $EB=BC=CE\Rightarrow \triangle BEC$ is equilateral $\Rightarrow\angle EBC=60^{\circ}\Rightarrow \angle CAE=30^{\circ}$ $\Rightarrow\angle BCE=60^{\circ}\Rightarrow \angle BDE=30^{\circ}$ Now $E' = E$, so $\angle BE'C=\angle BEC = 60^{\circ}\text{ }\square$

Let $\omega_1$ and $\omega_2$ be two circles centered at $B$ and $C$ with radius $BC$. $\omega_1$ and $\omega_2$ intersects at $E'$. $\angle EBC=\angle ECB=60^{\circ}\implies\angle EAC=\angle BDE=30^{\circ} $ because $A,C$ and $B,D$ pairwise lies on the same circle. Which means, $E\equiv E'$ and $\angle BEC=60^{\circ}$.