proof of Desargues’ theorem

The claim is that if triangles $A B C$ and $X Y Z$ are perspective froma point $P$ , then they are perspective from a line (meaning thatthe three points

AB\\cdot XY\\qquad BC\\cdot YZ\\qquad CA\\cdot ZX

are collinear) and conversely.

Since no three of $A, B, C, P$ are collinear, we can lay downhomogeneous coordinates such that

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

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

X=(1,p,p)\\qquad Y=(q,1,q)\\qquad Z=(r,r,1)

The equation for a line through $(x_{1},y_{1},z_{1})$ and $(x_{2},y_{2},z_{2})$ is

(y_{1}z_{2}-z_{1}y_{2})x+(z_{1}x_{2}-x_{1}z_{2})y+(x_{1}y_{2}-y_{1}x_{2})z=0\\;,

giving us equations for six lines:

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

whence

$\\displaystyle AB\\cdot XY$	$\\displaystyle=$	$\\displaystyle(pq-q,-pq+p,0)$
$\\displaystyle BC\\cdot YZ$	$\\displaystyle=$	$\\displaystyle(0,qr-r,-qr+q)$
$\\displaystyle CA\\cdot ZX$	$\\displaystyle=$	$\\displaystyle(-rp+r,0,rp-p).$

As claimed, these three points are collinear, since thedeterminant

\\left|\\begin{array}[]{ccc}pq-q&-pq+p&0\\\\0&qr-r&-qr+q\\\\-rp+r&0&rp-p\\end{array}\\right|

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

p(q-1)(r-1)x+(p-1)q(r-1)y+(p-1)(q-1)rz=0\\;.)

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