Let $a,b,c,d$ be the complex numbers corresponding to $A,B,C,D$ respectively. Assume the origin of the plane is $O$, the center of $(ABCD)$. Then $|a|=|b|=|c|=|d|\ (*)$. Let $g_a=\frac{b+c+d}3,g_b=\frac{a+c+d}3,g_c=\frac{a+b+d}3,g_d=\frac{a+b+c}3$ be the affixes of the centroids. We can then see that if $p=\frac{a+b+c+d}3$, then $|p-g_a|=|\frac a3|,|p-g_b|=|\frac b3|,|p-g_c|=|\frac c3|,|p-g_d|=|\frac d3|$, and combined with $(*)$ this yields $|p-g_a|=|p-g_b|=|p-g_c|=|p-g_d|$, so $G_a(g_a),G_b(g_b),G_c(g_c),G_d(g_d)$ lie on a circle having $P(p)$ as center.

orl wrote: Let $ABCD$ be a quadrilateral inscribed in a circle. Show that the centroids of triangles $ABC,$ $CDA,$ $BCD,$ $DAB$ lie on one circle. The problem is trivial! In fact, if A', B', C', D' are the centroids of triangles BCD, CDA, DAB, ABC, respectively, then we have to prove that these centroids A', B', C', D' lie on one circle. Now, if G is the centroid of all the four points A, B, C, D, then the mass-regrouping theorem (I guess the name is wrong) says that e. g. the point G lies on the segment AA', and divides this segment in the ratio AG : GA' = 3. Thus, GA' : AG = 1/3, so that GA' : GA = -1/3. This means that the homothety with center G and factor -1/3 maps the point A to the point A'. Similarly, the same homothety maps the points B, C, D to the points B', C', D'. And since the points A, B, C, D lie on one circle, it is clear that their images under this homothety must also lie on one circle, and thus we see that the points A', B', C', D' lie on one circle. Problem solved. Darij