Prove or disprove: For a set of at least 3 points not all collinear, we can always construct a triangle that contains the points, with the added condition that each of the triangle's edges has a point at its center.
Example: See the figure below. 7 points are scattered about at random, and the black triangle contains all of them. (Note that I consider the red point in the bottom left contained despite being intersected by the edge of the triangle.) Furthermore, the triangle's sides each have a point from the set at their center. I have colored these center points orange for convenient viewing, but there is nothing special about them: I might have selected another three center points when building a triangle to contain this set.