Parallel sides in regular polygons

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.

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

• Maybe considering the angles with the centroid of the regular polygon might help. – Arpan Apr 6 '15 at 3:51
• 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? – Greg Apr 6 '15 at 3:55
• 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. – Arpan Apr 6 '15 at 3:57
• 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... – Theo Apr 6 '15 at 5:11

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.