Suppose $XY$ intersects $\omega$ at points $P$ and $Q$, where $Q$ lies between $X$ and $Y$. We will show that $V$ and $W$ are the reflections of $A$ and $B$ with respect to the perpendicular bisector of $PQ$. From this, it follows that $AVWB$ is an isosceles trapezoid and hence $AB = VW$.

First, note that
$$\angle BZD = \angle AXY = \angle APQ + \angle BAP = \angle APQ + \angle BZP$$
so $\angle APQ = \angle PZV = \angle PQV$, and hence $V$ is the reflection of $A$ with respect to the perpendicular bisector of $PQ$.

Now, suppose $W'$ is the reflection of $B$ with respect to the perpendicular bisector of $PQ$, and let $Z'$ be the intersection of $YW'$ and $\omega$. It suffices to show that $B, X, D, Z'$ are concyclic. Note that
$$\angle YDC = \angle PDB = \angle PCB + \angle QPC = \angle W'PQ + \angle QPC = \angle W'PC = \angle YZ'C.$$
So $D, C, Y, Z'$ are concyclic. Next, $\angle BZ'D = \angle CZ'B - \angle CZ'D = 180^{\circ} - \angle BXD$ and due to the previous concyclicity we are done.

Alternative solution 1:
Using cyclic quadrilaterals $BXDZ$ and $ABZV$ in turn, we have $\angle ZDY = \angle ZBA = \angle ZCY$. So $ZDCY$ is cyclic.
Using cyclic quadrilaterals $ABZC$ and $ZDCY$ in turn, we have $\angle AZB = \angle ACB = \angle WZV$ (or $180^{\circ} - \angle WZV$ if $Z$ lies between $W$ and $C$).
So $AB = VW$ because they subtend equal (or supplementary) angles in $\omega$.

Alternative solution 2:
Using cyclic quadrilaterals $BXDZ$ and $ABZV$ in turn, we have $\angle ZDY = \angle ZBA = \angle ZCY$. So $ZDCY$ is cyclic.
Using cyclic quadrilaterals $BXDZ$ and $ABZV$ in turn, we have $\angle DXA = \angle VZB = 180^{\circ} - BAV$. So $XD \parallel AV$.
Using cyclic quadrilaterals $ZDCY$ and $BCWZ$ in turn, we have $\angle YDC = \angle YZC = \angle WBC$. So $XD \parallel BW$.
Hence $BW \parallel AV$ which implies that $AVWB$ is an isosceles trapezium with $AB = VW$.