Desargues' theorem

Desargues' theorem
Perspective triangles. Corresponding sides of the triangles, when extended, meet at points on a line called the axis of perspectivity. The lines which run through corresponding vertices on the triangles meet at a point called the center of perspectivity. Desargues' theorem guarantees that the truth of the first condition is necessary and sufficient for the truth of the second.

In projective geometry, Desargues' theorem, named in honor of Gérard Desargues, states:

Two triangles are in perspective axially if and only if they are in perspective centrally.

Denote the three vertices of one triangle by a, b, and c, and those of the other by A, B, and C. Axial perspectivity means that lines ab and AB meet in a point, lines ac and AC meet in a second point, and lines bc and BC meet in a third point, and that these three points all lie on a common line called the axis of perspectivity. Central perspectivity means that the three lines Aa, Bb, and Cc are concurrent, at a point called the center of perspectivity.

The result is true in the usual Euclidean plane but special care needs to be taken in exceptional cases, as when a pair of sides are parallel, so that their "point of intersection" recedes to infinity. Mathematically the most satisfying way of resolving the issue of exceptional cases is to "complete" the Euclidean plane to a projective plane by "adding" points at infinity following Poncelet.

Desargues's theorem is true for the real projective plane, for any projective space defined arithmetically from a field or division ring, for any projective space of dimension unequal to two, and for any projective space in which Pappus's theorem holds. However, there are some non-Desarguesian planes in which Desargues' theorem is false.


Projective versus affine spaces

In an affine space such as the Euclidean plane a similar statement is true, but only if one lists various exceptions involving parallel lines. Desargues' theorem is therefore one of the most basic of simple and intuitive geometric theorems whose natural home is in projective rather than affine space.


By definition, two triangles are perspective if and only if they are in perspective centrally (or, equivalently according to this theorem, in perspective axially). Note that perspective triangles need not be similar.

Under the standard duality of plane projective geometry (where points correspond to lines and collinearity of points corresponds to concurrency of lines), the statement of Desargues's theorem is self-dual:[1] axial perspectivity is translated into central perspectivity and vice versa. The Desargues configuration (below) is a self-dual configuration.[2]

Proof of Desargues' theorem

Desargues's theorem holds for projective space of any dimension over any field or division ring, and also holds for abstract projective spaces of dimension at least 3. In dimension 2 the planes for which it holds are called Desarguesian planes and are the same as the planes that can be given coordinates over a division ring. There are also many non-Desarguesian planes where Desargues's theorem does not hold.

Three-dimensional proof

Desargues's theorem is true for any projective space of dimension at least 3, and more generally for any projective space that can be embedded in a space of dimension at least 3.

Desargues' theorem can be stated as follows:

If A.a, B.b, C.c are concurrent, then
(A.B) ∩ (a.b), (A.C) ∩ (a.c), (B.C) ∩ (b.c) are collinear.

The points A, B, a, and b are coplanar because of the assumed concurrency of A.a and B.b. Therefore, the lines (A.B) and (a.b) belong to the same plane and must intersect. Further, if the two triangles lie on different planes, then the point (A.B) ∩ (a.b) belongs to both planes. By a symmetric argument, the points (A.C) ∩ (a.c) and (B.C) ∩ (b.c) also exist and belong to the planes of both triangles. Since these two planes intersect in more than one point, their intersection is a line that contains all three points.

This proves Desargues's theorem if the two triangles are not contained in the same plane. If they are in the same plane, Desargues's theorem can be proved by choosing a point not in the plane, using this to lift the triangles out of the plane so that the argument above works, and then projecting back into the plane. The last step of the proof fails if the projective space has dimension less than 3, as in this case it may not be possible to find a point outside the plane.

Monge's theorem also asserts that three points lie on a line, and has a proof using the same idea of considering it in three rather than two dimensions and writing the line as an intersection of two planes.

Two-dimensional proof

As there are non-Desarguesian projective planes in which Desargues' theorem is not true,[3] some extra conditions need to be met in order to prove it. These conditions usually take the form of assuming the existence of sufficiently many collineations of a certain type, which in turn leads to showing that the underlying algebraic coordinate system must be a division ring (skewfield).[4]

Relation to Pappus' theorem

Pappus's hexagon theorem states that, if a hexagon AbCaBc is drawn in such a way that vertices a, b, and c lie on a line and vertices A, B, and C lie on a second line, then each two opposite sides of the hexagon lie on two lines that meet in a point and the three points constructed in this way are collinear. A plane in which Pappus's theorem is universally true is called Pappian. Hessenberg (1905)[5] showed that Desargues's theorem can be deduced from three applications of Pappus's theorem (Coxeter 1969, 14.3).

The converse of this result is not true, that is, not all Desarguesian planes are Pappian. Satisfying Pappus's theorem universally is equivalent to having the underlying coordinate system be commutative. A plane defined over a non-commutative division ring (a division ring that is not a field) would therefore be Desarguesian but not Pappian. However, due to Wedderburn's theorem, which states that all finite division rings are fields, all finite Desarguesian planes are Pappian. There is no known, satisfactory geometric proof of this fact.

The Desargues configuration

The Desargues configuration viewed as a pair of mutually inscribed pentagons: each pentagon vertex lies on the line through one of the sides of the other pentagon.

The ten lines involved in Desargues' theorem (six sides of triangles, the three lines Aa, Bb, and Cc, and the axis of perspectivity) and the ten points involved (the six vertices, the three points of intersection on the axis of perspectivity, and the center of perspectivity) are so arranged that each of the ten lines passes through three of the ten points, and each of the ten points lies on three of the ten lines. Those ten points and ten lines make up the Desargues configuration, an example of a projective configuration. Although Desargues' theorem chooses different roles for these ten lines and points, the Desargues configuration itself is more symmetric: any of the ten points may be chosen to be the center of perspectivity, and that choice determines which six points will be the vertices of triangles and which line will be the axis of perspectivity. The Desargues configuration has a symmetry group of order 120; that is, there are 120 different ways of permuting the points and lines of the configuration in a way that preserves its point-line incidences. One way of constructing the same configuration in a way that makes these symmetries more readily apparent is to choose five planes in three-dimensional space, and to form the Desargues configuration as the set of ten points where three planes meet and the set of ten lines where two planes meet. Then, the 120 different permutations of these five planes each correspond to symmetries of the configuration.

As a projective configuration the Desargues configuration has the notation (103103), meaning that each of its ten points is incident to three lines and each of its ten lines is incident to three points. Its ten points can be viewed in a unique way as a pair of mutually inscribed pentagons, or as a self-inscribed decagon (Hilbert & Cohn-Vossen 1952). The Desargues graph, a 20-vertex bipartite symmetric cubic graph, is so called because it can be interpreted as the Levi graph of the Desargues configuration, with a vertex for each point and line of the configuration and an edge for every incident point-line pair.

A non-Desargues (103103) configuration.

There also exist eight other (103103) configurations (that is, sets of points and lines in the Euclidean plane with three lines per point and three points per line) that are not incidence-isomorphic to the Desargues configuration, one of which is shown at left. In all of these configurations, any chosen point has three other points that are not collinear with it. But in the Desargues configuration, these three points are always collinear with each other (if the chosen point is the center of perspectivity, then the three points form the axis of perspectivity) while in the other configuration shown in the illustration these three points form a triangle of three lines. As with the Desargues configuration, the other depicted configuration can be viewed as a pair of mutually inscribed pentagons.

See also


  1. ^ This is due to the modern way of writing the theorem. Historically, the theorem only read, "In a projective space, a pair of centrally perspective triangles is axially perspective" and the dual of this statement was called the converse of Desargues' theorem and was always referred to by that name. See (Coxeter 1964, pg. 19)
  2. ^ (Coxeter 1964) pp. 26–27.
  3. ^ The smallest examples of these can be found in Room & Kirkpatrick 1971.
  4. ^ (Albert & Sandler 1968), (Hughes & Piper 1973), and (Stevenson 1972).
  5. ^ According to (Dembowski 1968, pg. 159, footnote 1), Hessenberg's original proof is not complete; he disregarded the possibility that some additional incidences could occur in the Desargues configuration. A complete proof is provided by Cronheim 1953.


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External links

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