Let the usual five postulates of Euclid been given.

Let's take also this postulate: "If two points lies on the same plane, the whole straight line joining the two points lies on that plane".

Is it possible to prove that if two straight lines intersect each other, there exist a plane containing them both?

EDIT: I don't want to prove ULTRA-RIGOROUSLY in Hilbert-style (for example I don't question the existence of planes and I don't bother about uniqueness). I want to prove it as Euclid would prove it.

With this further postulate it's simple to prove these two facts:

If Two planes that meets in a point, meets each other in a straight line

Proof: Let $\alpha$ and $\beta$ the two given planes meeting in A. Take any points B and C on the plane $\alpha$, and not on the plane $\beta$ but on the same side of it. Join AB, AC and produce BA to F (on the opposite side of $\beta$). Join CF. Then since B and F are on opposite sides of the plane $\beta$, C and F are also on opposite sides of it. Therefore CF must meet the plane $\beta$ in a point G. Then since A, G are both in each of the planes $\alpha$ and $\beta$, the straight line AG is in both planes.

If two straight lines AB and AC meeting in A lies on the same plane, then the triangle ABC lies on the same plane.

Proof: Trivial.

EDIT 2: You proposed me counterexamples in 4 dimensions, solutions with algeabric geometry... etc. I repeated 1000 times that I would like to find a solution suitable in 300 BC!!! It's not to hard to understand that Galois Theory or Riemann surfaces cannot be used! xD

Here it is an example of solution that I DON'T LIKE by Robert Simson

Let any plane pass through the straight line AB (produced indefinitely) as axis until it passes through the point C. Then, since the points A, C are in this plane, the straight line AC is on the plane and so it is BC.

I don't like the using of rotating motion, but this is in fact a solution. I want to find alternative proofs.

  • $\begingroup$ No, I didn't intend what you say. I mean don't question that given a point there is a plane passing through it. $\endgroup$ – Benzio Dec 15 '14 at 19:31
  • $\begingroup$ OK, so that's what you intended. Thanks for clarifying! $\endgroup$ – rschwieb Dec 15 '14 at 20:03
  • $\begingroup$ In $4$ dimensions two planes can meet in a single point ($xy$ and $zw$ planes). How will you rule this out? $\endgroup$ – Aaron Meyerowitz Dec 16 '14 at 2:52
  • $\begingroup$ Edit 2 in the post $\endgroup$ – Benzio Dec 16 '14 at 11:08
  • $\begingroup$ Oh, you have only six axioms and you are speaking about triangles in the second theorem without even definig them... If you think that your second theorem is "trivial" you can prove anything you want writing "Proof: trivial". $\endgroup$ – ajotatxe Dec 16 '14 at 20:13

In book 11 Euclid addresses solid geometry. Here is a treatment of the result in question along with a comment that it is not that successful.

There are other, more scholarly, free on line public domain sources but the one I linked to was a pioneering dynamic Java project (explore!). It is hyperlinked and easy to explore .

That was book 11 proposition 2 where proposition 1 is "A part of a straight line cannot be in the plane of reference and a part in plane more elevated." About which Joyce says

The problem is that there are no postulates for solid geometry. The postulates in Book I apparently refer to an ambient plane. ..... Without any postulates for nonplanar geometry it is impossible for solid geometry to get off the ground.

It's been intensely thought about through the ages. Theon , for one is worth a look and the commentaries of Heath for perspective. Certain of Euclid's propositions are better seen as further axioms. The question is which things are best to take as axioms.

Euclid seems to assume that if points A,B are in a plane E so is the whole line AB. I'd think that

  • there are 4 points no 3 on a line and not in the same plane

  • if A,B,C are not colinear then there is a plane E containing all 3

would be good choices. Then one could derive that if F is another such it contains the same points as E.

But as I said, there are 2000 years of precedent.

  • $\begingroup$ ... that's the point! I know that proposition XI.2 is not convincing. So I want to find if there is a way to make a convincing proof with the tools Euclid had. $\endgroup$ – Benzio Dec 15 '14 at 19:33
  • $\begingroup$ And I just added my own "sixth postulate"... so now there is at least a postulate for solid geometry :-) $\endgroup$ – Benzio Dec 15 '14 at 19:34
  • $\begingroup$ I put an "Edit 2" to the post :-) $\endgroup$ – Benzio Dec 16 '14 at 11:13

There are several sources which provide a rigorous introductory axiomatic treatment of 3-dimensional geometry, including:

  1. The book "Foundations of Three-Dimensional Euclidean Geometry" by I.Vaisman.

  2. The book "Non-Euclidean Geometry", by H. S. M. Coxeter. (If you look hard enough, you might find a pdf or djvu file freely, alas illegally, available online.)

  3. "Ten Axioms for Three Dimensional Euclidean Geometry" by Lino Gutierrez Novoa Proceedings of the American Mathematical Society, Vol. 19, No. 1 (1968), pp. 146-152. (Available for free from JStor).

If you are a college student, you might find these in your university library or get them through the inter-library loan.

  1. If you just want a list of axioms, look here.

These treatments (1, 2 and 3) begin with certain sets of axioms and include a proof that the space described axiomatically is isomorphic to the 3-dimensional affine space equipped with the metric given by a positive definite quadratic form on the underlying 3-dimensional vector space. Once you have this, you can prove any of the statements you are interested in by appealing to basic linear algebra; of course, it is quite likely that these statement are already proven in the process of establishing this isomorphism. (For instance, you can just declare that your axioms are the axioms of the 3-dimensional real affine vector space equipped with the standard metric and a "plane" means an affine hyperplane, while a "line" means an affine line. Then the property you are asking about becomes a simple linear algebra exercise.)


Probably not. Your sixth postulate says: "IF there is a plane that contains two points...", but it doesn't guarantee the existence of any plane. And what you want to prove is the existence of a certain plane.

  • 2
    $\begingroup$ Why the donwvote? $\endgroup$ – ajotatxe Dec 15 '14 at 18:57
  • $\begingroup$ I edited the post $\endgroup$ – Benzio Dec 15 '14 at 19:08

Euklid's postulates talk about points, lines, and circles. Especially by the latter, we see that it is aimed an planar geometry. In fact, the parallel postulate (depending on its exact formulation) implies that two lines are either parallel or intersecting, hence at any rate in the same plane Consequently, there is only one plane. - If on the other hand you introduce "are on the same plane" as a new equivalence relation for points, the axioms allow that it is the identity relation, hence no two distinct points would be on the same plane ...


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