Desargues' theorem

Let A₁B₁C₁ and A₂B₂C₂ be two triangles. Consider two conditions:

1. Lines A₁A₂, B₂B₂, C₁C₂ joining the corresponding vertices are concurrent.
2. Points ab, bc, ca of intersection of the (extended) sides A₁B₁ and A₂B₂, B₁C₁ and B₂C₂, C₁A₁ and C₂A₂, respectively, are collinear.

Desargues' theorem claims that 1. implies 2. It's dual asserts that 1. follows from 2. In particular, the dual to Desragues' theorem coincides with its converse.

(In the applet below each of the triangles as well as each of the vertices is draggable.)

Two triangles that satisfy the first condition are said to be perspecitve from a point. Two triangles that satisfy the second condition are said to be perspecitve from a line. Desargues' theorem thus claims that two triangles perspective from a point are perspective from a line. Its dual asserts that two triangles perspective from a line are also perspective from a point.

Monge's theorem can be derived from that of Desargues and in fact is the latter in disguise. The existence of an orthic axis of a triangle is also an immediate consequence of Desargues' theorem. In a somewhat disguised form Desargues' theorem establishes a relationship between a triangle and a cevian triangle of a point not on a triangle itself.

Curiously, Desragues' theorem admits an intuitive proof if considered as a statement in the 3-dimensional space, but is not as easy in the 2-dimensional case, where it is often taken as an axiom.

Following is the proof (kindly supplied by Hubert Shutrick) that adopts the 3-dimensional perspective.

Proof of Desargues' Theorem

Let P be the common point of A₁A₂ , B₁B₂, C₁C₂ and suppose , in the first case, that C₁ is not in the plane PA₁B₁. Denote the line of intersection of the planes of the triangles by l. Since A₁A₂ and B₁B₂ meet at P, they are in the same plane which meets the line l in the point ab. In the same way ca and bc must be on l as required.

Suppose now that the two triangles are in the same plane. Choose a point E outside the plane and let C₁' be a point of the segment between E and C₁. The line EC₂ meets the line PC₁' in a point denoted C₂'. The proof above then applies to the triangles A₁B₁C₁' and A₂B₂C₂', but seen from E, that is, projecting all the points and lines from E onto the original plane, we get the result for the original triangles.

Proof of the converse

Once again, the three-dimensional case is easy. If the triangles are not in the same plane but in planes that intersect in l, then, since A₁B₁ and A₂B₂ intersect in ab, they are in the same plane which contains A₁A₂ and B₁B₂. Similarly, there is a plane containing A₁A₂ and C₁C₂ and one containing B₁B₂ and C₁C₂. These three planes intersect in the required point P.

In the case when the triangles are in the same plane, we pull them apart as in the previous proof. Choose a point E that is not in the plane and a point A₁' between E and A₁. Let B₁' be the intersection of EB₁ with abA₁' and, similarly, C₁' is the intersection of EC₁ with acA₁'. The line B₁'C₁' will meet l in bc as B₁C₁ does. The three-dimensional case then gives the point P' where A₁'A₂, B₁'B₂ and C₁'C₂ meet and EP' intersects the original plane in the required point P.

2D problems that benefit from a 3D outlook

1. 4 travellers
2. Desargues' Theorem
3. Soddy Circles and Eppstein's Points
4. Symmetries in a triangle
5. Three circles - #1
6. Three circles - #2
7. Three circles problem - #3

Desargues' Theorem

• Desargues' Theorem
• 2N-Wing Butterfly Problem
• Cevian Triangle
• Do You Speak Mathematics?
• Desargues in the Bride's Chair (with Pythagoras)
• Menelaus From Ceva
• Monge from Desargues
• Monge via Desargues
• Nobbs' Points, Gergonne Line
• Soddy Circles and David Eppstein's Centers
• Pascal Lines: Steiner and Kirkman Theorems II
• Pole and Polar with Respect to a Triangle
• The Lepidoptera of the Circles

Copyright © 1996-2008 Alexander Bogomolny