Without the parallel postulate, how can one prove that in any triangle $ABC$, $AB+AC>BC$?

Most often, we introduce the triangle inequality a priori as part of the definition of a metric, but here everything we have is the four basic postulates (save the parallel postulate) and the five principles in Euclid's Elements.

The proof Euclid himself made is actually pretty close, only that he invoked the proposition that the exterior angle is greater than an interior angle, which should be false without the parallel postulate (e.g. in spherical geometry).

Is it possible to completely avoid the parallel postulate and prove the triangle inequality?

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    $\begingroup$ See this. (Or google "Euclid proposition 20". In Euclid, the first 28 propositions are proven without using the parallel postulate.) $\endgroup$ – David Mitra Sep 13 '17 at 12:53
  • $\begingroup$ @DavidMitra this is where I came from. It traces back to the "greater side greater angle" proposition which again traces back to "exterior angle greater than interior angle" which inevitably builds on the parallel postulate. $\endgroup$ – Vim Sep 13 '17 at 12:56
  • $\begingroup$ So, you have qualms with Prop. 16 (exterior angle greater than int.)? $\endgroup$ – David Mitra Sep 13 '17 at 12:58
  • $\begingroup$ @DavidMitra exactly. And as the author of these notes said in his comment below, this proposition is indeed false without parallel postulate e.g. in elliptic geometry. $\endgroup$ – Vim Sep 13 '17 at 13:00
  • $\begingroup$ @DavidMitra thanks I'll check it out later. But if he really doesn't use PP to prove Proposition 16, then it should hold true regardless of whether PP is true or false. So I still don't quite buy into your claim that he didn't use PP altogether. $\endgroup$ – Vim Sep 13 '17 at 14:15

I already discussed this point several times (see this answer, for instance): Euclid's exterior angle theorem is perfectly valid in absolute geometry (i.e. euclidean geometry without parallel postulate), as its proof does not rely on the parallel postulate (and if you do not believe that, you should point out where this postulate is used in the proof).

Elliptic geometry IS NOT absolute geometry, because it also does away with Euclid's second postulate.


I understand what you are after and I am regretfully not conversant enough with the foundations of geometry to provide an rigorous answer.

Hoping not to waste anybody's time, I wonder though if the following elementary line of reasoning could be re-phrased to meet your requirements: I apologise in advance if it contains gross inaccuracies.

I believe that using the first four postulates one could prove that the time taken travelling at constant unitary speed between two distinct points is strictly positive: let us call this time $t$

So let us take a triangle $ABC$: without loss of generality let the length of the side $AB$ equal $1$. Now, if the triangular inequality is violated $$\vert BC \vert + \vert AC \vert \leq \vert AB \vert $$ Let us consider first the strict inequality $$\vert BC \vert + \vert AC \vert < \vert AB \vert $$

Let us define $$\gamma = \frac{\vert BC \vert + \vert AC \vert }{\vert AB \vert } $$ and from what is said, $\gamma < 1$.

If an ant had to travel at constant speed and wishes to minimize the travel time between $A$ and $B$, it will certainly choose the path $ACB$, and will travel it in time $\gamma$.

Now, if one builds similar triangles on the sides $BC$ and $AC$ it turns out the optimal path will take time $\gamma^2$ (here I worry I miss essential facts on non-euclidean geometries). Iterating one could prove that there exists a path, along which the travelling time is $\gamma^n $, so one could travel from $A$ to $B$ in a small as desired time, which cannot be right in any geometry.

We are left with the case whereby $$\vert BC \vert + \vert AC \vert = \vert AB \vert $$ which can be maybe be proven geometrically as follows (again, the "proof" might well be circular, as I might be giving for granted facts about geometry which actually stem from the triangular inequality itself).

Draw a circumference $C_1$ having $B$ as center and $AB$ as radius.

Draw a circumference $C_2$ having $C$ as center and $AC$ as radius.

The line, extension of the segment $AC$, will intersect the circumference in a point $D$. Now $$\vert DB \vert = \vert BC\vert + \vert CD\vert = \vert BC\vert + \vert CA\vert =\vert AB\vert $$ the last equality following from the (assumed) equality in the triangular inequality. This means $D$ should lie on $C_1$, which is not possible.

  • $\begingroup$ Thanks for your effort. I haven't read the entire answer but it seems that the opposition of the triangle inequality isn't $BC+AC\color{red}{<}AB$ as you said, but it should be less than or equal to instead. (In Riemannian manifolds though, there exists a unique point for the equality to hold, so this problem shouldn't be essential.) $\endgroup$ – Vim Sep 13 '17 at 12:49
  • $\begingroup$ Indeed, I will edit shortly. $\endgroup$ – An aedonist Sep 13 '17 at 12:52
  • $\begingroup$ I have completed reading your answer. The idea is great, but yeah, I agree with you that one should be wary about "similar triangles" in an noneuclidean setting. $\endgroup$ – Vim Sep 13 '17 at 12:59
  • $\begingroup$ @Vim, yes that bit is dodgey, but then, let renounce to similarity altogether. Just build any triangle: if the inequality is strict, we will still get that the travelling time goes as $\prod_1^n \gamma_i$, where each $\gamma_i < 1$, so really similarity is not essential. $\endgroup$ – An aedonist Sep 13 '17 at 13:22
  • $\begingroup$ how can you guarantee the triangle inequality is violated every time you build a new triangle? Also, even if we are guaranteed to obtain a sequence $\gamma_n<1$, it doesn't necessarily mean $\prod \gamma_n=0$, say, $\gamma_n=(1-1/n^2)$. $\endgroup$ – Vim Sep 13 '17 at 13:23

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