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Note that $\angle XYO=\angle XAO=\angle OAY=\angle OXY$, hence the triangle $△OXY$ is isosceles. Let $K,L,M,N$ be the midpoints of $AB,AC,AD,XY$ respectively. Note that $K,L,M$ are collinear because they lie on the midline of $△ABC$ parallel to $BC$. Moreover, $M$ lies on $XY$ by assumption. Since $XY\parallel BC\parallel KL$, the lines $KL$ and $XY$ do not coincide. This means that the lines $XY$ and $KL$ intersect at $M$.

Note that $K,L,N$ are the projections of $O$ onto $AX,AY$ and $XY$, respectively. Therefore $K,L,N$ lie on the Simson line of $O$.

This means that $M=N$, i.e. the segments $AD$ and $XY$ share a common midpoint. Therefore $AXDY$ is a parallelogram.