---
tag: 5Z19Y02MME
kind: statement
role: theorem
name: Desargues's theorem
msc: [51A30, 51A20]
uses: [ZT9A6JK8QF, DANFNN8C4Y, 0AFNNTRTJE]
about: [DANFNN8C4Y]
sources:
  - url: https://en.wikipedia.org/wiki/Desargues%27s_theorem
same_as:
  wikidata: Q841893
  wikipedia: Desargues's theorem
---

## Statement

Let $k$ be a field, or more generally a division ring, and work in the [[projective plane|DANFNN8C4Y]] $\mathbb{P}^2(k)$.
Let $A, B, C$ and $A', B', C'$ be two triangles, that is, two triples of non-[[collinear|0AFNNTRTJE]] points.
Assume $A \ne A'$, $B \ne B'$, $C \ne C'$, and that the corresponding sides $AB$ and $A'B'$, $BC$ and $B'C'$, $CA$ and $C'A'$ are distinct lines.
Then the lines $AA'$, $BB'$, $CC'$ pass through a common point if and only if the three points
$$AB \cap A'B', \quad BC \cap B'C', \quad CA \cap C'A'$$
are collinear.

## Remarks

The two triangles are then said to be in perspective from a point and in perspective from a line.
For two triangles in different planes of a [[projective space|ZT9A6JK8QF]] of dimension $3$, the proof is short: the three intersection points lie on both planes, hence on their common line.
The plane case follows by viewing the plane figure as a projection of a figure in space, or by a computation with [[homogeneous coordinates|FS9FRJ17D2]].
The converse direction is the dual statement, so it follows by the principle of [[duality|DCMP1A0138]].

In the Euclidean plane, if two triangles are homothetic, as $A' = 2A$, $B' = 2B$, $C' = 2C$ with centre the origin, the corresponding sides are parallel, so they meet on the line at infinity.

The theorem does not follow from the axioms of a projective plane alone.
It fails in the projective completion of the [[Moulton plane|3V9GYYXM1H]] and in other [[non-Desarguesian planes|HYE5FT4M34]].
It holds in a projective plane exactly when the plane is isomorphic to $\mathbb{P}^2(D)$ for some division ring $D$.

## Explanations

### L1 human

Draw two triangles so that the lines through matching corners all meet at one point, as if one triangle were the shadow of the other cast by a lamp.
Then extend matching sides until they cross: the three crossing points always lie on one straight line.
The statement only talks about points and lines, with no lengths or angles.

### L2 human

Desargues's theorem is the dividing line between planes that come from algebra and planes that do not.
A projective plane in which it holds can be coordinatized by a division ring.
It holds automatically in every projective space of dimension at least $3$, so the [[Veblen–Young theorem|NPNNXWMVCC]] needs no extra axiom there.
[[Pappus's hexagon theorem|B6C3J2GJWP]] implies it, by Hessenberg's theorem, and Pappus holds exactly when the coordinate division ring is commutative.
The little Desargues theorem, where the centre lies on the axis, characterizes [[Moufang planes|7F4HSNJJH8]].