Let A1B1C1 and A2B2C2 be two triangles. Consider two conditions:
- Lines A1A2, B2B2, C1C2 joining the corresponding vertices are concurrent.
- 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. (The proof fails in some exceptional configurations, e.g., when A1, A2, B1, and B2 are collinear. These are simple enough to be treated individually and will not be considered here. We simply restrict the theorem to two triangles in general position, i.e., assuming that A1≠A2, B1≠B2, C1≠C2 and that, similarly, no two corresponding side lines coincide.)
Proof of Desargues' Theorem
Let O be the common point of A1A2, B1B2, C1C2 and suppose , in the first case, that C1 is not in the plane OA1B1. Denote the line of intersection of the planes of the triangles by l. Since A1A2 and B1B2 meet at O, 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 OC1' 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.
(Obviously, the proof falls flat lifting C1 and C2 if A1, B1, A2, B2, are collinear because the new triangles are again coplanar. This can always be avoided by lifting one of the other corresponding pairs.)
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 O.
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 O' where A1'A2, B1'B2 and C1'C2 meet and EO' intersects the original plane in the required point O.
Proof from Menelaus' Theorem
For convenience, the points of intersection of the corresponding sides will be renamed to
|B1P / PC1||·||C1C2 / C2O||·||OB2 / B2B1||= -1|
|C1Q / QA1||·||A1A2 / A2O||·||OC2 / C2C1||= -1|
|A1R / RB1||·||B1B2 / B2O||·||OA2 / A2A1||= -1|
Multiplying the three equalities we obtain after simplification
|B1P / PC1||·||C1Q / QA1||·||A1R / RB1||= -1|
which, by the converse of Menelaus' theorem, implies the collinearity of the three points P, Q, R.
2D Problems That Benefit from a 3D Outlook
- Four Travellers, Solution
- Desargues' Theorem
- Soddy Circles and Eppstein's Points
- Symmetries in a Triangle
- Three Circles and Common Chords
- Three Circles and Common Tangents
- Three Equal Circles
- Menelaus from 3D
- Stereographic Projection and Inversion
- Stereographic Projection and Radical Axes
- Sum of Squares in Equilateral Triangle
- 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
- Desargues' Hexagon
- The Lepidoptera of the Circles
Copyright © 1996-2016 Alexander Bogomolny