proof of Desargues’ theorem

The claim is that if triangles ABC and XYZ are perspective from a point P, then they are perspective from a line (meaning that the three points


are collinearMathworldPlanetmath) and conversely.

Since no three of A,B,C,P are collinear, we can lay down homogeneous coordinatesMathworldPlanetmath such that

A=(1,0,0)  B=(0,1,0)  C=(0,0,1)  P=(1,1,1)

By hypothesis, there are scalars p,q,r such that

X=(1,p,p)  Y=(q,1,q)  Z=(r,r,1)

The equation for a line through (x1,y1,z1) and (x2,y2,z2) is


giving us equations for six lines:

AB : z=0
BC : x=0
CA : y=0
XY : (pq-p)x+(pq-q)y+(1-pq)z=0
YZ : (1-qr)x+(qr-q)y+(qr-r)z=0
ZX : (rp-p)x+(1-rp)y+(rp-r)z=0


ABXY = (pq-q,-pq+p,0)
BCYZ = (0,qr-r,-qr+q)
CAZX = (-rp+r,0,rp-p).

As claimed, these three points are collinear, since the determinant


is zero. (More precisely, all three points are on the line


Since the hypotheses are self-dual, the converse is true also, by the principle of duality.

Title proof of Desargues’ theorem
Canonical name ProofOfDesarguesTheorem
Date of creation 2013-03-22 13:47:51
Last modified on 2013-03-22 13:47:51
Owner drini (3)
Last modified by drini (3)
Numerical id 5
Author drini (3)
Entry type Proof
Classification msc 51A30