Theorem of Desargues

The theorem of Desargues is one of (if not it) the more important theorem of the projective Géométrie. It binds two Triangle S and the right S which they determine. It is related to the concept of harmonic groups .

In projective geometry

In projective geometry, its statement is the following:

Are ABC and A' B' It two triangles without common point such as lines (AA'), (BB') (CC') are convergent then points of intersection of the right-hand sides

(AB) and (A' B')
(AC) and (A' It)
(BC) and (B' It)

are located on the same line.

If no point of intersection is an unsuitable point, one obtains the configuration opposite: The dual of this theorem gives also the reciprocal one:

ABC and A' B' It two triangles without common point such as the points of intersection of the right-hand sides

(AB) and (A' B')
(AC) and (A' It)
(BC) and (B' It)

are located on the same line

then lines (AA'), (BB') (CC') are convergent (in a clean or unsuitable point).

Of projective geometry, a simple demonstration consists in using a Projective application which transforms the triangles (ABC) and (A' B' It) in ( $A_1B_1C_1$ ) and ( $A'_1B'_1C'_1$ ) so that the points of intersection of ( $A_1B_1$ ) and ( $A'_1B'_1$ ) and of ( $A_1C_1$ ) and ( $A'_1C'_1$ ) are unsuitable. The lines ( $A_1B_1$ ) and ( $A'_1B'_1$ ) are then parallel as well as the lines ( $A_1C_1$ ) and ( $A'_1C'_1$ ). One then returned to the preceding configuration which proves that ( $B_1C_1$ ) and ( $B'_1C'_1$ ) are parallel and that their point of intersection is thus unsuitable (located on the same line as the two precedents). One from of then deduced the property on the lines from departure.

In geometry " classique" , this form of the theorem of Desargues can be shown by imagining that our drawing represents in prospect a tetrahedron SABC cut by a plan (A' B' It). Lines (AB) and (A' B') being in plan (SAB) their apparent point of intersection is a point of intersection real common to both plans (ABC) and (A' B' It), it is the same of the point of intersection of right-hand sides (BC) and (B' It) and of the point of intersection of the right-hand sides (AC) and (A' It). The plans (ABC) and (A' B' It) are secant according to a line (d) which contains these three points of intersection.

In geometry closely connected

In geometry closely connected, the theorem of Desargues must be supplemented to take into account the particular cases due to the parallelism. The Arguesian Géométrie makes it possible to reveal these particular statements like corollaries of the general case:

If ABC and A' B' They is two triangles, without common point such as

(AB) // (A' B')
(BC) // (B' It)
(CA)/(It A')

then lines (AA'), (BB') and (CC') are convergent or parallel.

In the first configuration, one supposes that two of the right-hand sides (for example (AA') and (BB')) are secant in S. the proof is carried out by building the point M on (SC) such as (A' M)/(AC). The Théorème from Thalès makes it possible to prove that then (MB')/(CB). However there exists one point located, on the one hand on the parallel to (AC) carried out by A', on the other hand on the parallel to (BC) carried out by B' and this point is It. Thus M and It are confused and points SCC' are thus aligned.

In the second configuration, one finds no line secant: they are thus parallel.

Reciprocal : Are ABC and A' B' It two triangles without common point such as

lines (AA'), (BB') and (CC') are convergent or parallel
(AB)/(A' B')
(AC)/(A' It)

then (BC)/(B' It).

With convergent lines, we find a consequence of the theorem of Thalès. With parallel straight lines, the demonstration uses the existence of parallelogram (AA' B' B) and (AA' It C) to deduce the existence from it from parallelogram (BB' It C).

Theorem of Desargues and axiomatic definition

At the beginning of XXme century, David Hilbert tries an axiomatic definition of the plan closely connected. On a space provided with points, it defines lines as whole of points checking certain axioms from which

By two points distinct passes one and only one line.
Any line contains at least two points. There exist at least three not aligned points.
By a point has external on a line (d), it passes only one straight line parallel with (d) (parallel I. E. does not have common point with).

He discovers whereas these axioms do not make it possible to show the theorem of Desargues. One needs the existence of the space or the axiom of the Congruence. In other words, this statement has value of axiom in geometry (closely connected or projective) plane.

In 1902, Moulton exhibe an example of plan - the plan of Moulton - checking the three preceding axioms and in which the theorem of Desargues is false.

Bonds

Girard Desargues, its life, its work.
the Theorem of Hessenberg which establishes the link between the Théorème of Pappus and this theorem of Desargues.
the Arguesian Géométrie makes it possible to establish the link between the two versions of this theorem.

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