# 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

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

are collinear) and conversely.

Since no three of $A,B,C,P$ are collinear, we can lay down homogeneous 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 the determinant

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

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, by the principle of duality.

Title proof of Desargues’ theorem ProofOfDesarguesTheorem 2013-03-22 13:47:51 2013-03-22 13:47:51 drini (3) drini (3) 5 drini (3) Proof msc 51A30