Let $M$ be the midpoint of $AB$ and $Z=PX\cap QY$. We know that $CM=AM=BM=\frac{1}{2}AB$ so $ACGP$ and $BCGQ$ are isosceles trapezoids with $GP\parallel AC$ and $GQ\parallel BC$, therefore $GC=AP=BQ=\frac{2}{3}CM=\frac{1}{3}AB\rightarrow GM=PM=QM=\frac{1}{6}AB$. We observe that $GPZQ$ is a rectangle and since $M$ is the midpoint of $PQ$ then $CM$ passes through $Z$. From the cyclic $CGPX$ we have that $\angle XCG=\angle XPG=90$, in the same way we conclude that $\angle YCG=90$. Thus the points $X,C,Y$ are collinear and $XYZ$ is a right triangle with height $ZC=ZM+CM=\frac{1}{6}AB+\frac{1}{2}AB=\frac{2}{3}AB$. We know that $CX*CY=ZC^2=\frac{4}{9}AB^2$, so $\frac{CX*CY}{AB^2}=\frac{\frac{4}{9} AB^2}{AB^2}=\frac{4}{9}$

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