Problem 952

11. For the given quadrilateral $ABCD$ (Fig. 21), we retain the same notations for centroids as in the previous problem: $M_{2}$ and $M_{4}$ are the centroids of triangles $ACD$ and $ABC$ respectively. Let $A_{1}$ and $A_{2}$ be the points of division on sides $AB$ and $AD$, closest to vertex $A$; $B_{1}$ and $B_{2}$ be the points of division on sides $BA$ and $BC$, closest to vertex $B$; $C_{1}$ and $C_{2}$ be the points of division on sides $CB$ and $CD$, closest to vertex $C$; and finally, $D_{1}$ and $D_{2}$ be the points of division on sides $DC$ and $DA$, closest to vertex $D$. Since $(B_{1}B_{2}) \parallel (A_{1}C_{1}) \parallel (AC)$ and $(D_{1}D_{2}) \parallel (A_{2}C_{2}) \parallel (AC)$, it follows that $(D_{1}D_{2}) \parallel (B_{1}B_{2})$. Similarly, we obtain that $(A_{1}A_{2}) \parallel (C_{1}C_{2})$. Thus, the four lines $A_{1}A_{2}$, $B_{1}B_{2}$, $C_{1}C_{2}$, and $D_{1}D_{2}$, intersecting, form a parallelogram. The centroid $M_{2}$ of triangle $ACD$ is located at the midpoint of segment $A_{2}C_{2}$. The centroid $M_{4}$ of triangle $ABC$ is located at the midpoint of segment $A_{1}C_{1}$. The centroid of quadrilateral $ABCD$ lies on the line $M_{2}M_{4}$, which passes through the center of the parallelogram, since $(M_{2}M_{4})$ coincides with the axis of the strip between the lines $A_{1}A_{2}$ and $C_{1}C_{2}$. Similarly, it can be shown that the line $M_{1}M_{3}$, where $M_{1}$ and $M_{3}$ are the centroids of triangles $ABD$ and $BCD$ respectively, also passes through the center of the parallelogram. Therefore, the centroid of the quadrilateral coincides with the center of the constructed parallelogram.