One of the more interesting things we can say in neutral geometry (that is, without assuming the parallel postulate, but assuming e.g. the rest of Hilbert's axioms) is the following:

Suppose that a single triangle whose angles sum to $180^\circ$ exists. Then the sum of the angles in every triangle is $180^\circ$. (And therefore the parallel postulate holds.)

Here is the outline of a proof. First, define the defect of a triangle to be $180^\circ$ minus the sum of its angles. It can be proved in neutral geometry that the defect is nonnegative, and that it's additive: if a triangle is chopped up into many smaller triangles, its defect is the sum of the smaller triangles' defects.

In particular, if one triangle is contained inside another, its defect is less than or equal to that of the large triangle.

We start with a triangle $\triangle ABC$ such that $\angle A + \angle B + \angle C = 180^\circ$. Simply by copying segments and angles (and the SAS triangle congruence axiom) we can tile the plane with copies of $\triangle ABC$. For any other triangle $\triangle XYZ$, we find a large triangle $\triangle A'B'C'$ which is made up of many copies of $\triangle ABC$ (and has the same angles), but contains $\triangle XYZ$. Then the defect of $\triangle XYZ$ is less than or equal to the defect of $\triangle A'B'C$, which is $0$.

The incomplete step here is "tile the plane". We have a way to generate a finite tiling with arbitrarily many copies of $\triangle ABC$ in all directions; we need to show that we can extend it to cover an arbitrary point in the plane.

This feels like an application of Archimedes's axiom. But how exactly do we get there using Archimedes?

  • $\begingroup$ It seems to me that you don't really need to tile the whole plane, just enough of the plane to contain a triangle $A'B'C'$ similar to the original $ABC$ and big enough to contain $XYZ$. $\endgroup$ – Andreas Blass Jul 14 '18 at 1:01
  • $\begingroup$ True. We can, for instance, build a copy of $\triangle XYZ$ with $X$ moved to $A$ and $Y$ on line $AB$, and try to make the tiling big enough to cover it. We would still need to cover the arbitrary point $Z$ in that case, so I'm not sure our life gets any easier. $\endgroup$ – Misha Lavrov Jul 14 '18 at 13:27

This is the second theorem of Legendre. Indeed, Archimedes axiom is necessary for this proof. You can find a proof for example in the book of Franz Rothe pages 242-246 (http://math2.uncc.edu/~frothe/3181all.pdf).

  • $\begingroup$ I'm familiar with the argument using Saccheri quadrilaterals, but I would prefer not to go that route or even to define what a Saccheri quadrilateral is. If Archimedes's axiom can be used to complete my argument, I think it would be a more intuitive justification. $\endgroup$ – Misha Lavrov Jul 15 '18 at 0:03

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