The bisector of $\angle BAC$ and the perpendicular bisector of $BC$ meet at $P$, the midpoint of the minor arc $\widehat{BC}$ (they are different lines as $AB \neq AC$). In particular $OP$ is perpendicular to $BC$ and intersects it at $M$, the midpoint of $BC$.

Denote by $Y'$ the reflection of $Y$ with respect to $OP$. Since $\angle BYC = \angle BY'C$, it suffices to prove that $BX CY'$ is cyclic.

We have
$$\angle XAP = \angle OPA = \angle EYP.$$
The first equality holds because $OA = OP$, and the second one because $EY$ and $OP$ are both perpendicular to $BC$ and hence parallel. But $\{Y, Y'\}$ and $\{E, D\}$ are pairs of symmetric points with respect to $OP$, it follows that $\angle EYP = \angle DY'P$ and hence
$$\angle XAP = \angle DY'P = \angle XY'P.$$
The last equation implies that $XAY'P$ is cyclic. By the powers of $D$ with respect to the circles $(XAY'P)$ and $(ABPC)$ we obtain
$$XD \cdot DY' = AD \cdot DP = BD \cdot DC$$
It follows that $BX CY'$ is cyclic, as desired.