So I've noticed a couple of things about regular polygons with an even number of sides but I'm having a hard time proving them, these are all very obvious, and I think perhaps induction is the best way to prove them for any (even) n:

  1. The opposite sides in a regular polygon are parallel.
  2. Number the vertices: {1,2....2n}, if you take the side that goes from say V1 to V2, the diagonals that skip an even number of vertices, i.e., V2nV3, V2n-1V4, etc... are parallel to the given side (perhaps this can be phrased better).
  3. The diagonals that go from one vertex to the opposite one are concurrent.

It's quite clear in the images: https://en.wikipedia.org/wiki/Regular_polygon#/media/File:Regular_polygon_6_annotated.svg https://en.wikipedia.org/wiki/Regular_polygon#/media/File:Regular_polygon_12_annotated.svg

I can't seem to find this anywhere, maybe because it's too obvious to even mention it, but still thanks for any help.

  • $\begingroup$ Maybe considering the angles with the centroid of the regular polygon might help. $\endgroup$
    – Arpan
    Apr 6, 2015 at 3:51
  • $\begingroup$ For number 3 yes, but the parallel lines I'm talking about in 1 and 2 do not necesarilly pass through the centroid. Or how do you mean? $\endgroup$
    – Greg
    Apr 6, 2015 at 3:55
  • $\begingroup$ I'm not entirely sure if it will help or not. But maybe connect the vertices to the centroid... You'll get angle subtended by each side at the center = $\frac{2\pi}{2n}$. Maybe that could lead you somewhere. $\endgroup$
    – Arpan
    Apr 6, 2015 at 3:57
  • $\begingroup$ Actually, the first one is true for all equiangular polygons (en.m.wikipedia.org/wiki/Equiangular_polygon) (but with an even number of sides, of course). To prove that, just consider the angle that each side forms with the $x$ axis; you'll see that this doesn't depend on the length of the side... $\endgroup$ Apr 6, 2015 at 5:11

1 Answer 1


This is evidently independent of scale and rotation. So you might as well treat the vertices as $$ v_j = (\cos (\frac{2\pi j}{2n}), \sin (\frac{2\pi j}{2n})) $$ or, in complex variables terms, $$ v_j = \exp(\frac{2\pi \mathbf i j}{2n}) $$ Once you do that, your claims should all be pretty straightforward consequences of the algebra.


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