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My question concerns the classical Desargues Theorem and its simplest version

The small Desargues Theorem: Let $A$, $B$, $C$ be three distinct parallel lines and $a,a'\in A$, $b,b'\in B$, $c,c'\in C$, be points such that $ab\parallel a'b'$ and $bc\parallel b'c'$. Then $ac\parallel a'c'$.

The small Desargues Theorem does not involve any Projective Geometry, so could have been known to ancient geometers (Thales, Euclides, Proclus, Pappus etc.).

Question. Is there any evidence that the small Desargues Theorem was known to ancient Greeks?

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    $\begingroup$ May be more appropriate on the History of Science and Mathematics Stack Exchange. $\endgroup$
    – Brian Hopkins
    Commented Jul 14, 2023 at 15:37
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    $\begingroup$ @BrianHopkins Thank you for the suggestion. I have copied my question there: hsm.stackexchange.com/questions/15567/… $\endgroup$
    – Taras Banakh
    Commented Jul 14, 2023 at 15:51
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    $\begingroup$ When you ask "Is it true or not?", it reads to me sort of like you are asking whether or not the theorem is true (which of course it is, and you know it is). Would you be amenable to changing the wording to "Was it?" or "Was it known?" (Or of course you could just delete that sentence, since it is asked explicitly in the following demarcated text.) $\endgroup$
    – LSpice
    Commented Jul 17, 2023 at 16:21
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    $\begingroup$ The attempt to close this post is unreasonable. Maybe we should look into MO accepting an official guideline that whatever research mathematicians are interested in, is considered par for the course at MO. @TarasBanakh $\endgroup$
    – Mikhail Katz
    Commented Jul 17, 2023 at 16:40
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    $\begingroup$ This question would not do better at the HSM site because answering it requires a level of mathematical expertise not possessed by most users there. $\endgroup$
    – Mikhail Katz
    Commented Jul 18, 2023 at 9:09

3 Answers 3

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This is more of a comment than an answer, but I think it deserves to be stated explicitly:

I don't know if the Greeks knew this result, but lest there be any doubt that the Greeks could have proved it, here is a proof using essentially only proposition 34 in book I of Euclid's Elements, which states that

two opposite sides of a parallelogram are equal to one another

(a “parallelogram” here means a quadrilateral whose opposite sides are parallel, and “equal” means having equal length).

In the present situation, $aa'b'b$ is a parallelogram and $bb'c'c$ is a parallelogram, and our goal is to show that $aa'c'c$ is.

Now the quoted proposition I.34 implies that $aa' = bb'$ and that $bb' = cc'$ as distances, so (as things equal to a third are equal to each other, this is Euclid's first “axiom”) $aa' = cc'$. Furthermore, the line $A=(aa')$ and $B=(bb')$ are parallel, and so are the lines $B=(bb')$ and $C=(cc')$, so $A$ and $C$ are parallel (this is proposition I.30 in Euclid). Now $aa'c'c$ has the two opposite sides $aa'$ and $cc'$ that are parallel and of equal length, so by the converse of the aforementioned proposition, $aa'c'c$ is a parallelogram.

OK, I couldn't find the converse of I.34 explicitly in Euclid, but it is easy to deduce it from I.34 itself: let $c^*$ be the point such that $aa'c^*c$ is a parallelogram (i.e., the intersection of the line $C$ with the parallel to $(ac)$ through $a'$), applying I.34 again we see that $cc^*=aa'$ and we have noted that $cc'=aa'$, so $cc^* = cc'$ and since they are on the same line¹, $c'=c^*$. (I suspect that the Greeks would have done this in a different way, but I have absolutely no doubt that the converse of I.34 was clear to them.)

  1. Astute readers will note that $c^*$ could conceivably be on the other side of $c$ from $c'$ from what I said, but this kind of omissions, merely supported by a figure, abound in Euclid. Pretty much the same reasoning is in I.35 (except that areas are used instead of distances).

Addendum: To summarize, I am tempted to say that the result being asked about (which again is a particular affine form of what is often known as the “little Desargues” theorem) is pretty much Euclid's proposition I.34: basically, we are to show that $\vec{aa'} = \vec{bb'}$ and $\vec{bb'} = \vec{cc'}$ implies $\vec{aa'} = \vec{cc'}$ (where “$\vec{xx'} = \vec{yy'}$” means by definition “$xx'y'y$ is a parallelogram”), and even though Euclid didn't know about vectors, he knew about lengths and parallels, and the above proof states that the lengths and directions are the same.

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  • $\begingroup$ Thank you for your answer. I also have a simple (and standard synthetic) proof of the small Desargues Theorem using Proposition 15 and 16 from the Euclid's book XI (go out to the 3-d dimension and then project back). $\endgroup$
    – Taras Banakh
    Commented Jul 17, 2023 at 16:13
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    $\begingroup$ There's a dichotomy between the two proofs, which was basically noted by Hilbert. You can prove the theorem by using congruences or by passing to 3D. $\endgroup$
    – arsmath
    Commented Jul 17, 2023 at 18:36
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In fact, I asked this question hoping to find some appropriate name that can be legally used for naming affine spaces satisfying the small Desargues Theorem. Now I have learned that in Projective Geometry, projective planes satisfying the projective analog of the small Desargues Theorem are called Moufang planes, which was also mentioned in the comment of @Gro-Tsen. So, it is logical to call affine planes satisfying the small Desargues Theorem Moufang affine planes. Is such a choice of terminology sufficiently justified, or there is a better one?

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    $\begingroup$ For what it's worth, I think this is a perfectly sensible terminology (at least if I am correct in my belief that this is precisely the structure of a projective Moufang plane minus a line qualified as “line at infinity”). I decline to comment, however, on whether “Moufang affine plane” or “affine Moufang plane” is preferable (or whether the adjectives commute). $\endgroup$
    – Gro-Tsen
    Commented Jul 24, 2023 at 15:10
  • $\begingroup$ @Gro-Tsen The adjectives "Moufang" and "affine" do not commute because I have a notion of an affine space (an incidence geometry in which the Playfair axiom of Parallel holds) and "Moufang" is a partial case of affine spaces in which the small Desargues Theorem holds. So, all affine spaces of dimension $\ge 3$ are Desarguesian and Moufang and some affine spaces of dimension 2 (i.e., affine planes) are Moufang, but there are also non-Moufang planes, for example, the Moulton plane, see en.wikipedia.org/wiki/Moulton_plane $\endgroup$
    – Taras Banakh
    Commented Jul 24, 2023 at 16:50
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Your "small Desargues theorem" is weaker than the full Desargues theorem (as pointed out by Gro-Tsen in a comment). Thus the small Desargues theorem cannot be proved in any given non-Mounfang plane. So to give a synthetic proof (since they didn't present analytic proofs), the Greeks would have had to go out to 3 dimensions. I think this makes it unlikely that they would have known the result.

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    $\begingroup$ The theorem as stated by OP is a particular affine form of the special case of the Desargues theorem whereby the center of perspectivity lies on the axis of perspectivity. This theorem is known to hold in Moufang planes (e.g., the octonionic projective plane) which may fail to be Desarguian. So it is strictly weaker than the full Desargues theorem. $\endgroup$
    – Gro-Tsen
    Commented Jul 17, 2023 at 15:22
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    $\begingroup$ @MikhailKatz But ancient Greeks knew stereometry very well. What is the problem for Greeks to go out the plane to the 3-d dimension? $\endgroup$
    – Taras Banakh
    Commented Jul 17, 2023 at 15:33
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    $\begingroup$ The full Desargues theorem is a theorem in Euclidean geometry (though to avoid using projective geometry you have to consider a bunch of cases), so it must follow from the order properties of the reals. So there's no requirement to pass to 3d to prove the special case here. $\endgroup$
    – arsmath
    Commented Jul 17, 2023 at 15:44
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    $\begingroup$ @MikhailKatz The proof of small Desargues Theorem uses only two facts: (1) a plane $\Pi_1$ is parallel to a plane $\Pi_2$ if $\Pi_1$ contains two intersecting lines that are parallel to the plane $\Pi_2$ and (2) intersecting two parallel planes by a third plane, we obtain two parallel lines. Those two results were certainly known to Euclid. More precisely, they are Propositions 15 and 16 in the book XI of Euclid, see aleph0.clarku.edu/~djoyce/java/elements/bookXI/bookXI.html $\endgroup$
    – Taras Banakh
    Commented Jul 17, 2023 at 15:57
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    $\begingroup$ @TarasBanakh, That's very interesting. This wasn't clear from your "question" at all. I think you should include this information. So are you saying that all the tools were there to provide a synthetic proof? $\endgroup$
    – Mikhail Katz
    Commented Jul 17, 2023 at 15:59

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