Ok, I'll have a crack at it.

On second thought, you need to also know some axioms such as the relationship between a regular polygons area and perimeter.

So that is: kl^{2} = k2l

k is a constant depending on the number of sides, and l is a side length.

We see that Kl^{2} = (pi)r^{2} in this case.

Therefore, we set K = pi and l = r; If you want to make l = diameter, then k = (pi)/2

Substitute this in for the other equation (transitive property, I think. Three years since geometry and proofs) and you get 2(pi)r = C.

I think I just made up some rules, but feel free to approve of this. Then again, you may need some calculus to really prove it. Calculus which I don't have.

EDIT: This basic rule should work (the axiom that I outlined) with regular polygons. It works with rectangles (k = 1), triangles, though, is on shaky footing. But I know there is a formal relation somewhere.

EDITEDIT: triangles k is sqr(3)/4