Let $O$ be the centre of $(ABCD)$. Let $H_A, H_B, H_C, H_D$ and $G_A, G_B, G_C, G_D$ be the orthocentres and centroids of $\triangle DAB$, $\triangle ABC$, $\triangle BCD$ and $\triangle CDA$ respectively. Let $M$ be the midpoint of $AB$. Recall that $O$ lies on $H_A G_A$, etc., with $OG_A : G_A H_A = 1 : 2$. Also, we have $DG_A : G_A M = 2 : 1 = CG_B : G_B M$. From these, we know that $H_A H_B \parallel G_A G_B \parallel DC$, and
$$H_A H_B = 3 G_A G_B = CD.$$

By symmetry, we have $H_B H_C = DA$, $H_C H_D = AB$, $H_D H_A = BC$, $H_A H_C = CA$ and $H_B H_D = DB$. Thus, $H_A H_B H_C H_D$ is congruent to $CDAB$. It follows

that $H_A H_B H_C H_D$ is concyclic. Now, note that $G_A G_B G_C G_D$ is the image of $H_A H_B H_C H_D$ under the homothety with centre $O$ and ratio $\frac{1}{3}$. Therefore, $G_A G_B G_C G_D$ is cyclic.