Proving the concurrence of the three lines formed by connecting midpoints of opposite sides of a hexagon

Let's consider a convex hexagon where we know the midpoints of each side.

I'm currently trying to show that if we connect the midpoints of the opposite sides into lines that the three lines will all intersect at one common point. Now because this was in the area unit of my work, I figured that it would have to relate to area in some way or another. And so far, I've been able to find that if you take any one of those three lines that connect the midpoints on the opposite sides of a hexagon, it will divide the hexagon into two pentagons, and these pentagons have an equal area to each other. But I'm not sure how to relate this information back to the problem. If anyone can help out, that would be create.

Honestly, this takes almost no effort to find a counterexample. And if even this figure is not sufficient proof, we can choose an explicit set of vertices; e.g.,

$$A = (12,0) \\ B = (10, 10) \\ C = (-11, 3) \\ D = (-12, 0) \\ E = (-11, -3) \\ F = (10, -10).$$ Then compute the midpoints, and then show that the lines joining the opposite midpoints do not share a common intersection.

• Thanks! I was too focused in on trying to prove it, and I really must have just skipped over finding a counterexample. Sep 20 '20 at 1:34
• @Edwards Geogebra or Geometer's Sketchpad are very useful tools for sketching figures. Sep 20 '20 at 1:37

Here's a counterexample: The red dots are the vertices, and the blue lines connect the midpoints of the sides.

• Would that be counted as a convex hexagon? I thought that convex polygons need to have angles strictly less than 180 degrees. Sep 20 '20 at 1:03
• @Edwards I don't know what the convention is regarding whether to count $180^\circ$ angles, but the vertices could be moved by an infinitesimal amount to make the angles strictly less than $180^\circ$ while not moving the lines significantly, so the counterexample still holds. Sep 20 '20 at 1:16