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# Desargues

[dey-zarg, Eng. dey-zahrg]
Desargues, Gérard, 1591-1661, French mathematician and engineer, a founder of modern geometry. He discovered the theorems on involutions and transversals known by his name and worked on conic sections. His writings, lost for a time, were republished in 1864. His purely mathematical texts, in French, ed. by René Taton, were republished in 1951.
In projective geometry, Desargues' theorem, named in honor of Gérard Desargues, states:

In a projective space, two triangles are in perspective axially if and only if they are in perspective centrally.

To understand this, denote the three vertices of one triangle by (lower-case) a, b, and c, and those of the other by (capital) A, B, and C. Axial perspectivity is the condition satisfied if and only if the point of intersection of ab with AB, and that of intersection of ac with AC, and that of intersection of bc with BC, are collinear, on a line called the axis of perspectivity. Central perspectivity is the condition satisfied if and only if the three lines Aa, Bb, and Cc are concurrent, at a point called the center of perspectivity.

## Projective versus affine spaces

In an affine space a similar statement is true only if one lists various exceptions involving accidentally parallel lines. Desargues' theorem is therefore one of the most basic of simple and intuitive geometric theorems whose natural home is in projective rather than affine space.

## Self-duality

By definition, two triangles are perspective if and only if they are in perspective axially (or, equivalently according to this theorem, in perspective centrally). Note that perspective triangles need not be similar.

Under the standard duality of plane projective geometry (where points correspond to lines and collinearity of points corresponds to concurrency of lines), the statement of Desargues's theorem is self-dual: axial perspectivity is translated into central perspectivity and vice versa.

## The Desargues configuration

The ten lines involved (six sides of triangles, the three lines Aa, Bb, and Cc, and the axis of perspectivity) and the ten points involved (the six vertices, the three points of intersection on the axis of perspectivity, and the center of perspectivity) are so arranged that each of the ten lines passes through three of the ten points, and each of the ten points lies on three of the ten lines. Those ten points and ten lines make up the Desargues configuration (see projective configuration). (It is an amusing exercise to show that those incidence conditions can also be satisfied by a configuration of ten points and ten lines that is not incidence-isomorphic to the Desargues configuration.) The statement of the theorem above may misleadingly connote that the Desargues configuration has less symmetry than it really has: Any of the ten points may be chosen to be the center of perspectivity, and that choice determines which six points will be the vertices of triangles and which line will be the axis of perspectivity.

As a projective configuration the Desargues configuration has the notation (103103). Its ten points can be viewed in a unique way as a pair of mutually inscribed pentagons, or as a self-inscribed decagon (Hilbert and Cohn-Vossen 1952).

## Proof of Desargues' theorem

The truth of Desargues' theorem in the plane is more readily deduced by modeling it in 3-dimensional space and subsequently projecting the result into the plane than by actually constructing the proof in 2-space. Two triangles cannot be in perspective unless they fit into a space of dimension 3 or less; thus in higher dimensions the affine span of the two triangles is always a subspace of dimension no higher than 3.

Desargues' theorem can be stated as follows:

If A.a, B.b, C.c are concurrent, then
(A.B)∩(a.b), (A.C)∩(a.c), (B.C)∩(b.c) are collinear.

In purely symbolic terms, using the cross product and the dot product, Desargues' theorem can be stated like so: If

$\left(A times a\right) cdot \left(B times b\right) times \left(C times c\right) = 0$

then

$\left(\left(A times B\right) times \left(a times b\right)\right) cdot \left(\left(\left(A times C\right) cdot \left(a times c\right)\right) times \left(\left(B times C\right) times \left(b times c\right)\right)\right) = 0.$

Letting <X, Y, Z> denote the scalar triple product, Desargues' theorem can be stated thus: If

$langle A times a, B times b, C times crangle = 0$

then

$langle \left(A times B\right) times \left(a times b\right), \left(A times C\right) times \left(a times c\right), \left(B times C\right) times \left(b times c\right)rangle = 0.$

### First restatement

Knowing that a vector triple product
$X times \left(Y times Z\right)$
is equal to
$Y \left(X cdot Z\right) - Z \left(X cdot Y\right),$
one can derive the formula
$\left(X times Y\right) times \left(Z times W\right) = langle X,Y,Wrangle Z - langle X,Y,Zrangle W.$
From this last formula, one can further derive the identity
$langle U times V, W times X, Y times Zrangle = langle W,X,Zrangle langle U,V,Yrangle - langle W,X,Yrangle langle U,V,Zrangle.$

Through application of this identity, Desargues' theorem can be restated as follows:
If

$langle B,b,crangle langle A,a,Crangle = langle B,b,Crangle langle A,a,crangle$
then
$langle A times C, a times c, b times crangle langle A times B, a times b, B times Crangle = langle A times C, a times c, B times Crangle langle A times B, a times b, b times crangle.$

### Second restatement

Applying the identity again to the consequent of the first restatement of Desargues' theorem, commuting triple products, and cyclically permuting the vectors of each triple product, one obtains this second restatement:
If
$langle A,a,crangle langle b,B,Crangle = langle a,A,Crangle langle B,b,crangle$
then
$langle C,a,crangle langle b,A,Brangle = langle c,A,Crangle langle B,a,brangle.$

Notice that the left side of the consequent can be obtained from the left side of the antecedent through the substitutions AC, BA, CB. Also, the right side of the consequent can be obtained from the right side of the antecedent thought the substitutions ac, ba, cb.

### Third restatement

A theorem of vector calculus states that the product of two scalar triple products is equal to a determinant of a matrix whose elements are dot products determined by the rule
$M_\left\{ij\right\} = u_i cdot v_j, qquad langle u_1,u_2,u_3rangle langle v_1,v_2,v_3rangle = |M|.$

Applying this theorem to the second restatement yields this third one:
If

$left| begin\left\{matrix\right\} A cdot b & a cdot b & c cdot b A cdot B & a cdot B & c cdot B A cdot C & a cdot C & c cdot C end\left\{matrix\right\} right| = left| begin\left\{matrix\right\} a cdot B & A cdot B & C cdot B a cdot b & A cdot b & C cdot b a cdot c & A cdot c & C cdot c end\left\{matrix\right\} right|$
then
$left| begin\left\{matrix\right\} C cdot b & a cdot b & c cdot b C cdot A & a cdot A & c cdot A C cdot B & a cdot B & c cdot B end\left\{matrix\right\} right| = left| begin\left\{matrix\right\} c cdot B & A cdot B & C cdot B c cdot a & A cdot a & C cdot a c cdot b & A cdot b & C cdot b end\left\{matrix\right\} right|.$

### Fourth restatement

Expanding the determinants of the third restatement yields this fourth one:

If

$\left(A cdot b\right) \left(a cdot B\right) \left(c cdot C\right) + \left(a cdot b\right) \left(c cdot B\right) \left(A cdot C\right) + \left(c cdot b\right) \left(A cdot B\right) \left(a cdot C\right)$

$- \left(A cdot b\right) \left(c cdot B\right) \left(a cdot C\right) - \left(a cdot b\right) \left(A cdot B\right) \left(c cdot C\right) - \left(c cdot b\right) \left(a cdot B\right) \left(A cdot C\right)$

$= \left(a cdot B\right) \left(A cdot b\right) \left(C cdot c\right) + \left(A cdot B\right) \left(C cdot b\right) \left(a cdot c\right) + \left(C cdot B\right) \left(a cdot b\right) \left(A cdot c\right)$

$- \left(a cdot B\right) \left(C cdot b\right) \left(A cdot c\right) - \left(A cdot B\right) \left(a cdot b\right) \left(C cdot c\right) - \left(C cdot B\right) \left(A cdot b\right) \left(a cdot c\right)$

then

$\left(C cdot b\right) \left(a cdot A\right) \left(c cdot B\right) + \left(a cdot b\right) \left(c cdot A\right) \left(C cdot B\right) + \left(c cdot b\right) \left(C cdot A\right) \left(a cdot B\right)$

$- \left(C cdot b\right) \left(c cdot A\right) \left(a cdot B\right) - \left(a cdot b\right) \left(C cdot A\right) \left(c cdot B\right) - \left(c cdot b\right) \left(a cdot A\right) \left(C cdot B\right)$

$= \left(c cdot B\right) \left(A cdot a\right) \left(C cdot b\right) + \left(A cdot B\right) \left(C cdot a\right) \left(c cdot b\right) + \left(C cdot B\right) \left(c cdot a\right) \left(A cdot b\right)$

$- \left(c cdot B\right) \left(C cdot a\right) \left(A cdot b\right) - \left(A cdot B\right) \left(c cdot a\right) \left(C cdot b\right) - \left(C cdot B\right) \left(A cdot a\right) \left(c cdot b\right).$

### Fifth restatement

The first and fifth terms of each side of both equations (antecedent and consequent) of the fourth restatement end up being cancelled out, yielding this fifth restatement:

If

$\left(A cdot C\right) \left(B cdot c\right) \left(a cdot b\right) + \left(A cdot B\right) \left(C cdot a\right) \left(b cdot c\right)$

$- \left(A cdot b\right) \left(B cdot c\right) \left(C cdot a\right) - \left(A cdot C\right) \left(B cdot a\right) \left(b cdot c\right)$

$= \left(A cdot B\right) \left(C cdot b\right) \left(a cdot c\right) + \left(A cdot c\right) \left(B cdot C\right) \left(a cdot b\right)$

$- \left(A cdot c\right) \left(B cdot a\right) \left(C cdot b\right) - \left(A cdot b\right) \left(B cdot C\right) \left(a cdot c\right)$

then

$\left(A cdot c\right) \left(B cdot C\right) \left(a cdot b\right) + \left(A cdot C\right) \left(B cdot a\right) \left(b cdot c\right)$

$- \left(A cdot c\right) \left(B cdot a\right) \left(C cdot b\right) - \left(A cdot C\right) \left(B cdot c\right) \left(a cdot b\right)$

$= \left(A cdot B\right) \left(C cdot a\right) \left(b cdot c\right) + \left(A cdot b\right) \left(B cdot C\right) \left(a cdot c\right)$

$- \left(A cdot b\right) \left(B cdot c\right) \left(C cdot a\right) - \left(A cdot B\right) \left(C cdot b\right) \left(a cdot c\right).$

### Sixth restatement

Between the two equations of the fifth restatement there are eight different terms: each one showing up twice. Let the terms be relabeled as follows:
$t_1 = \left(A cdot C\right) \left(B cdot c\right) \left(a cdot b\right),$
$t_2 = \left(A cdot B\right) \left(C cdot a\right) \left(b cdot c\right),$
$t_3 = \left(A cdot b\right) \left(B cdot c\right) \left(C cdot a\right),$
$t_4 = \left(A cdot C\right) \left(B cdot a\right) \left(b cdot c\right),$
$t_5 = \left(A cdot B\right) \left(C cdot b\right) \left(a cdot c\right),$
$t_6 = \left(A cdot c\right) \left(B cdot C\right) \left(a cdot b\right),$
$t_7 = \left(A cdot c\right) \left(B cdot a\right) \left(C cdot b\right),$
$t_8 = \left(A cdot b\right) \left(B cdot C\right) \left(a cdot c\right).$
Then the fifth restatement becomes the following:
If
$t_1 + t_2 - t_3 - t_4 = t_5 + t_6 - t_7 - t_8$
then
$t_6 + t_4 - t_7 - t_1 = t_2 + t_8 - t_3 - t_5.$

### Seventh restatement

Move the terms on the right side of the antecedent's equation of the sixth restatement to the left side, and the terms on the left side of the consequent's equation to the right side. The result is:
If
$t_1 + t_2 - t_3 - t_4 - t_5 - t_6 + t_7 + t_8 = 0$
then
$0 = t_1 + t_2 - t_3 - t_4 - t_5 - t_6 + t_7 + t_8.$

The consequent is now seen to be identical to the antecedent, so that Desargues' theorem is seen to be true. Q.E.D.