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- 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 iff 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 iff 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 nothing similar is true unless 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.
By definition, two triangles are perspective iff 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. (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.
Proof of Desargues' theorem
The truth of Desargues' theorem in the plane is most readily deduced by getting it as a corollary to its truth in a 3-dimensional space rather than the 2-dimensional plane. 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.
Letting <X, Y, Z> denote the scalar triple product, Desargues' theorem can be stated thus: If
Knowing that a vector triple product
is equal to
one can derive the formula
From this last formula, one can futher derive the identity
Through application of this identity, Desargues' theorem can be restated as follows:
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:
Notice that the left side of the consequent can be obtained from the left side of the antecedent through the substitutions A→C, B→A, C→B. Also, the right side of the consequent can be obtained from the right side of the antecedent throught the substitutions a→c, b→a, c→b.
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
Applying this theorem to the second restatement yields this third one:
Expanding the determinants of the third restatement yields this fourth one:
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:
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:
Then the fifth restatement becomes the following:
- t1 + t2 - t3 - t4 = t5 + t6 - t7 - t8
- t6 + t4 - t7 - t1 = t2 + t8 - t3 - t5.
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:
- t1 + t2 - t3 - t4 - t5 - t6 + t7 + t8 = 0
- 0 = t1 + t2 - t3 - t4 - t5 - t6 + t7 + t8.
The consequent is now seen to be identical to the antecedent, so that Desargues' theorem is seen to be true. Q.E.D.
- Monge & d'Alembert Three Circles Theorem I - Dynamic Geometry by Antonio Gutierrez from Geometry Step by Step from the Land of the Incas.
- Monge & d'Alembert Three Circles Theorem II - Dynamic Geometry by Antonio Gutierrez from Geometry Step by Step from the Land of the Incas.
- Desargues' Theorem
- Monge via Desargues
- Monge via Desargues II
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