Pappus's hexagon theorem

In mathematics, Pappus's hexagon theorem (attributed to Pappus of Alexandria) states that It holds in a projective plane over any field, but fails for projective planes over any noncommutative division ring.

If one considers a pappian plane containing a hexagon as just described but with sides

is the line at infinity), one gets the affine version of Pappus's theorem shown in the second diagram.

have a point in common, one gets the so-called little version of Pappus's theorem.

[2] The dual of this incidence theorem states that given one set of concurrent lines

(Concurrent means that the lines pass through one point.)

Pappus's theorem is a special case of Pascal's theorem for a conic—the limiting case when the conic degenerates into 2 straight lines.

In general, the Pappus line does not pass through the point of intersection of

If the affine form of the statement can be proven, then the projective form of Pappus's theorem is proven, as the extension of a pappian plane to a projective plane is unique.

Because of the parallelity in an affine plane one has to distinct two cases:

The key for a simple proof is the possibility for introducing a "suitable" coordinate system: Case 1: The lines

The proof above also shows that for Pappus's theorem to hold for a projective space over a division ring it is both sufficient and necessary that the division ring is a (commutative) field.

[4][5] In general, Pappus's theorem holds for some projective plane if and only if it is a projective plane over a commutative field.

In that case an alternative proof can be provided, for example, using a different projective reference.

Because of the principle of duality for projective planes the dual theorem of Pappus is true: If 6 lines

If in the affine version of the dual "little theorem" point

The Thomsen figure plays an essential role coordinatising an axiomatic defined projective plane.

But there exists a simple direct proof, too: Because the statement of Thomsen's theorem (the closure of the figure) uses only the terms connect, intersect and parallel, the statement is affinely invariant, and one can introduce coordinates such that

[10] These are Lemmas XII, XIII, XV, and XVII in the part of Book VII consisting of lemmas to the first of the three books of Euclid's Porisms.

The lemmas are proved in terms of what today is known as the cross ratio of four collinear points.

The first of these, Lemma III, has the diagram below (which uses Pappus's lettering, with G for Γ, D for Δ, J for Θ, and L for Λ).

Here three concurrent straight lines, AB, AG, and AD, are crossed by two lines, JB and JE, which concur at J.

Then These proportions might be written today as equations:[11] The last compound ratio (namely JD : GD & BG : JB) is what is known today as the cross ratio of the collinear points J, G, D, and B in that order; it is denoted today by (J, G; D, B).

So we have shown that this is independent of the choice of the particular straight line JD that crosses the three straight lines that concur at A.

In particular It does not matter on which side of A the straight line JE falls.

Pappus does not explicitly prove this; but Lemma X is a converse, namely that if these two cross ratios are the same, and the straight lines BE and DH cross at A, then the points G, A, and Z must be collinear.

What we showed originally can be written as (J, ∞; K, L) = (J, G; D, B), with ∞ taking the place of the (nonexistent) intersection of JK and AG.

Pappus shows this, in effect, in Lemma XI, whose diagram, however, has different lettering: What Pappus shows is DE.ZH : EZ.HD :: GB : BE, which we may write as The diagram for Lemma XII is: The diagram for Lemma XIII is the same, but BA and DG, extended, meet at N. In any case, considering straight lines through G as cut by the three straight lines through A, (and accepting that equations of cross ratios remain valid after permutation of the entries,) we have by Lemma III or XI Considering straight lines through D as cut by the three straight lines through B, we have Thus (E, H; J, G) = (E, K; D, L), so by Lemma X, the points H, M, and K are collinear.

That is, the points of intersection of the pairs of opposite sides of the hexagon ADEGBZ are collinear.

That is, the points of intersection of the pairs of opposite sides of the hexagon BEKHZG are collinear.