APMO

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 $W^{\prime }$ is the reflection of $B$ with respect to the perpendicular bisector of $PQ$, and let $Z^{\prime }$ be the intersection of $Y{W}^{\prime }$ and $\omega$. It suffices to show that $B,X,D,Z^{\prime }$ are concyclic. Note that $$\angle YDC=\angle PDB=\angle PCB+\angle QPC=\angle W^{\prime }PQ+\angle QPC=\angle W^{\prime }PC=\angle Y{Z}^{\prime }C.$$ So $D,C,Y,Z^{\prime }$ are concyclic. Next, $\angle B{Z}^{\prime }D=\angle C{Z}^{\prime }B-\angle C{Z}^{\prime }D=180^{\circ }-\angle BXD$ 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 $AB=VW$.