Quadrilaterals; why
If we consider a quadrilateral whose vertices have abscissa \(x_0\), \(x_1\), \(x_2\) and \(x_3\), we get a new quadrilateral with vertices with abscissa \(x'_0\), \(x'_1\), \(x'_2\) and \(x'_3\), given by the identities \(x'_i=\frac{x_i+x_{i+1}}{2}\) for \(i\in\{0,1,2,3\}\) where \(x_4=x_0\).
The abscissa of the midpoint of the diagonal that links the vertices with abscissa \(x'_0\) and \(x'_2\) is given by \[\frac{x'_0+x'_2}{2}\;=\;\frac{1}{2}\left( \frac{x_0+x_1}{2}+\frac{x_2+x_3}{2}\right)\;=\; \frac{x_0+x_1+x_2+x_3}{4} \] while the abscissa of the midpoint of the diagonal that links the vertices with abscissa \(x'_1\) and \(x'_3\) is given by \[\frac{x'_1+x'_3}{2}\;=\;\frac{1}{2}\left( \frac{x_1+x_2}{2}+\frac{x_3+x_0}{2}\right)\;=\; \frac{x_0+x_1+x_2+x_3}{4}. \] This means that the diagonals of the new quadrilateral intersect at the midpoint of both. Its coordinates are given by the arithmetic mean of each of the respective coordinates of the four vertices of the initial quadrilateral (this point, called the center of gravity, or centroid, is the same for any of the quadrilaterals of the produced sequence of polygons). Thus, using criteria of equality of triangles, it is easy to conclude that the opposite sides are equal and the quadrilateral is a parallelogram. Moreover, the initial quadrilateral does not univocally determine the new parallelograms, that is, the same sequence of parallelograms can be obtained from different initial quadrilaterals. In fact, there are a multitude of different initial quadrilaterals giving rise to the same sequence of parallelograms! This situation is completely opposite to the previous one, in which the initial triangle univocally determined the sequence of triangles and in which the new triangles were all similar to the initial one.