Let $\angle BAC = 2\alpha$ and $\angle DBA = \theta$. Note that since $CE = BE = DE$, $E$ is the circumcenter of $\triangle DCB$ and therefore $\angle DEB = 2\angle ACB = 180^{\circ}- 2\alpha = \angle DAB$, so $DAEB$ is cyclic. Since $\angle ZEA = \frac{1}{2} \angle DEB + \theta = 90^{\circ} +\theta - \alpha$ and $\angle ZAE$ is right, we have that $\angle AZE = \alpha - \theta$, and $\angle ADE = \angle ABE = \alpha - \theta$ too, so $ADZE$ is cyclic, as desired.

Diagram stolem from @Sumgato Claim 1(AEDB is cyclic).We first notice that $AE \perp BC \implies AE$ passes through the mid point of $BC$(let this be called $U$), since $\triangle ABC$ is isoceles. We know that the perpendicular bisector of $BD$ passes throught the mid point of $BD$(Let this be called $V$). $\angle{EUB} = \angle{EVB} = 90^{\circ} \implies EUBV$ is cyclic. Since $\overline{UV}$ is the midsegment of $\triangle BDC$, $\implies \measuredangle{BDC} = \measuredangle{BVU} = \measuredangle{BEU} = \measuredangle{BEA} \implies AEDB$ is cyclic. Now same as Sumgato sar did $\angle{AEZ} = 90^{\circ} - \angle{AZE} = \angle{DBC} = \angle{DBA} + \angle{ABC} = \angle{DEA} + \angle{ACB} = \angle{DEA} + \angle{DAZ} \implies \angle{ADE} = 180^{\circ} - \angle{DAZ} - 90^{\circ} - \angle{DEA} = 90^{\circ} - (90^{\circ} - \angle{AZE}) = \angle{AZE} \implies ADEZ$ is cyclic.