From $\varangle F A H=\varangle F G H=\varangle F G D=\frac{1}{2} \varangle F A D=90^{\circ}-\varangle A F D$ we deduce that $A H \perp$ $D F$. Similarly, $\varangle D A I=180^{\circ}-\varangle D E I=180^{\circ}-\varangle D E F=\varangle D G F=\frac{1}{2} \varangle D A F$, so we also have $A I \perp D F$. Therefore, points $A, H, I$ are collinear. Analogously, we find that the triples of points $(A, K, J),\left(A, H^{\prime}, I^{\prime}\right)$ and $\left(A, K^{\prime}, J^{\prime}\right)$ are collinear.
Quadrilateral $H I J K$ is cyclic because $\varangle A I K=\varangle A D K=\varangle A G H=\varangle A J H$. Analogously, quadrilateral $H^{\prime} I^{\prime} J^{\prime} K^{\prime}$ is cyclic.
Finally, since $\varangle H^{\prime} J H=\varangle H^{\prime} G H=\varangle E G D=\varangle E F D=\varangle J F J^{\prime}=\varangle J H J^{\prime}$, quadrilateral $H J J^{\prime} H^{\prime}$ is an isosceles trapezoid with $H J \| H^{\prime} J^{\prime}$, so the perpendicular bisectors of $H J$ and $H^{\prime} J^{\prime}$ coincide. Analogously, the perpendicular bisectors of $I K$ and $I^{\prime} K^{\prime}$ coincide. Therefore the circumcenters of $H I J K$ and $H^{\prime} I^{\prime} J^{\prime} K^{\prime}$ coincide.
!