Disappointing It's enough to show $\{A, O, X, D, E\}$ are concyclic. Observe immediately that $O\in (AEX)$ and $O\in (ADX)$ and since $$\angle EAD=\angle EXD$$$\{A, E, D, X\}$ are concyclic and the result follows.

Sniped \[\angle ACE = CAE = \angle DAB = \angle DAB \]\[\angle AEX = 2\angle XCA = \angle BDA = 180 - 2\angle DBA \Longrightarrow AEXC \text{ is cyclic }.\]We know that $OE$ and $OE$ are the perpendicular bisectors of $AB$ and $AC$. \[\angle AEO = \angle AEX + \angle OEZ = 90 - \angle EAC.\]\[\angle AXO = 90 - \tfrac{\angle AOX}{2} = 90 - \angle EAC.\]Hence $O$ lies on $(AEXD)$ which directly implies the result.

An earlier (and harder) version of the problem was as follows: Triangle $ABC$ has circumcircle $\Gamma$. A variable point $X\neq A,B,C$ lies on $\Gamma$. Let $D$ be a point on line $BX$ such that $ AD = BD$. Let $E$ be a point on line $CX$ such that $ AE = CE$. Prove that, as $X$ varies, the circumcentre of $\triangle DEX$ moves along a fixed line that is independent of $X$.