Let be $\Gamma$ he circumference of diameter $AB$ and let be $G=DE\cap \Gamma$ then $GB=CG$ since $ED$ is perpendicular bisector. Since $AB$ is diameter $\angle AGB=90$ and let $F=DE\cap BC$ then $\angle BFD=90$ but $\angle GDB=\angle GAB$ and as $\angle AGB=90$ then $\angle CBD=\angle GBA$ therefore $\angle GBA=\angle GDA=\angle ACE$ by angles in the circumcircle of $\triangle ADC$ but by $DB=DC$ we have that $\angle BCD=\angle DBC=\angle GBA=\angle ACE$ and this completes the proof.