Desargues' theorem
Table of Contents
Introduction
Desargues' theorem says the following:
Two triangles are in perspectives axially if and only if they are in perspective centrally.
Let's discuss and visualize the two directions of the theorem.
We start with the triangle \(\Delta_1 = A_1 B_1 C_1\) and a point \(P\) in the plane (the situation is shown in the graphic below). From \(P\) we draw rays through the vertices of \(\Delta_1\). Picking a point on each vertex we obtain a new triangle \(\Delta_2 = A_2 B_2 C_2\) that is in central perspective with \(\Delta_1\). Now consider the pair of lines \(A_1 B_1\) and \(A_2 B_2\) corresponding to one of the sides of \(\Delta_1\) and \(\Delta_2\) respectively. They meet in the point \(X_{AB}\) shown in red. Similarly, the two other pairs of lines corresponding to the remaining sides of the triangles meet in the points \(X_{AC}\) and \(X_{BC}\). The "if" direction of Desargues' theorem tell us that the three red points lie on the same line.
For the other direction consider again the triangle \(\Delta_1 = A_1 B_1 C_1\) and a line \(L\) (this situation is shown in the next graphic). The lines through the sides of the triangle intersect \(L\) in three points. For each of the these points we pick another line that goes through that point. These three lines intersect pairwise and form the triangle \(\Delta_2 = A_2 B_2 C_2\) that is in axial perspective with \(\Delta_2\). The "only if" direction of the theorem tell us that the lines \(A_1 A_2\), \(B_1 B_2\) and \(C_1 C_2\) meet in the single point \(P\).
One can ask what happens if some or all of the sides of the triangles \(\Delta_1\) and \(\Delta_2\) are parallel. Then the theorem is still true projectively and some or all of the intersection points \(X_{AB}\), \(X_{BC}\) and \(X_{AC}\) lie on the line at infinity of the projective plane.
Proofs of Desargues' theorem
There are two proofs of Desargues' theorem: one of them considers the plane as embedded in a three-dimensional space, the other one works purely in the plane.
Space proof
We think of the pictures above as obtained by a projection of a three-dimensional configuration of points, lines and planes to a plane. In particular, we imagine the point \(P\) as one of the vertices of a simplex and the two triangles as obtained by slicing it with two different planes.
Because the lines \(\langle A_1, A_2\rangle\), \(\langle B_1, B_2\rangle\) and \(\langle C_1, C_2\rangle\) meet in the point \(P\), the four points \(A_1\), \(A_2\), \(B_1\) and \(B_2\) lie on the same plane. Therefore the lines \(\langle A_1, B_1\rangle\) and \(\langle A_2, B_2\rangle\) intersect in a point, \(X_{AB}\). Similarly, the lines \(\langle A_1, C_1\rangle\) and \(\langle A_2, C_2\rangle\) intersect in the point \(X_{AC}\) and finally \(\langle B_1, C_1\rangle\) intersects \(\langle B_2, C_2\rangle\) in \(X_{BC}\).
Consider now the two planes \(\alpha_1 = \langle A_1, B_1, C_1\rangle\) and \(\alpha_2 = \langle A_2, B_2, C_2\rangle\) in which the two triangles lie. Because the lines \(\langle A_1, B_1\rangle\) and \(\langle A_2, B_2\rangle\) are fully contained in \(\alpha_1\) and \(\alpha_2\) respectively, their intersection \(X_{AB}\) belongs to both \(\alpha_1\) and \(\alpha_2\). Similarly, the other intersection points \(X_{AC}\) and \(X_{BC}\) also lie on both planes. As two planes either intersect in a straight line or have no point in common it follows that the points \(X_{AB}\), \(X_{AC}\) and \(X_{BC}\) are on a line.
For the other direction, let the three points \(X_{AB}\), \(X_{AC}\) and \(X_{BC}\) be on a line \(L\). Then the two triangles lie in a two planes that intersect in \(L\). As the lines \(\langle A_1, B_1\rangle\) and \(\langle A_2, B_2\rangle\) meet in a point on \(L\), the four points \(A_1\), \(A_2\), \(B_1\) and \(B_2\) lie on the same plane, \(\beta_{AB}\). Similarly, the four points \(B_1\), \(B_2\), \(C_1\) and \(C_2\) lie on the same plane \(\beta_{BC}\) and the points \(A_1\), \(A_2\), \(C_1\) and \(C_2\) lie on the plane \(\beta_{AC}\).
Because the lines \(\langle A_1, A_2\rangle\) and \(\langle B_1, B_2\rangle\) are on \(\beta_{AB}\), they meet in a point \(P\). The line \(\langle A_1, A_2\rangle\) also meets the plane \(\beta_{BC}\) in \(P\), because \(\langle B_1, B_2\rangle\) is contained in \(\beta_{BC}\) and a plane and a line meet in a single point. Similarly, the line \(\langle B_1, B_2\rangle\) meets the plane \(\beta_{AC}\) in \(P\). Therefore \(P\) also belongs to the intersection of \(\beta_{BC}\) and \(\beta_{AC}\) which is the line \(\langle C_1, C_2\rangle\).
This proof uses Hilbert's axioms about incidence (group I), order (group II) and the parallel postulate (group III). It does not require the axioms about congruence.
Plane proof
It is not possible to prove Desargues' theorem purely in two dimensions without adding additional axioms about the congruence of triangles (group IV in Hilbert's numbering).
Menelaus's theorem says the following:
Consider a triangle with vertices \(A\), \(B\) and \(C\). Let \(D\) be a point on the line \(\langle A, B\rangle\), \(E\) a point \(\langle B, C\rangle\) and \(F\) a point on \(\langle A, C\rangle\). Then the points \(D\), \(E\) and \(F\) are collinear if and only if
\begin{align} \left|\frac{AF}{FC}\right| \left|\frac{CE}{EB}\right| \left|\frac{BD}{DA}\right| = 1. \end{align}
The theorem can be proven in two dimensions using similar triangles.
In the Desargues' configuration above, consider the the triangle formed by \(P\), \(A_1\) and \(C_1\). Then Menelaus tells us that
\begin{align} \left|\frac{P A_2}{A_2 A_1}\right| \left|\frac{A_1 X_{AC}}{X_{AC} C_1}\right| \left|\frac{C_1 C_2}{C_2 P}\right| = 1 \end{align}because the points \(X_{AC}\), \(C_2\) and \(A_2\) are collinear. Similarly, the triangle with vertices \(P\), \(B_1\) and \(C_1\) gives us
\begin{align} \left|\frac{P B_2}{B_2 B_1}\right| \left|\frac{B_1 X_{BC}}{X_{BC} C_1}\right| \left|\frac{C_1 C_2}{C_2 P}\right| = 1 \end{align}and finally the triangle \(P\), \(A_1\), \(B_1\) yields the equation
\begin{align} \left|\frac{P B_2}{B_2 B_1}\right| \left|\frac{B_1 X_{AB}}{X_{AB} A_1}\right| \left|\frac{A_1 A_2}{A_2 P}\right| = 1. \end{align}Multiplying the first and the last and dividing by the second equation we get
\begin{align} \left|\frac{B_1 X_{AB}}{X_{AB} A_1}\right| \left|\frac{A_1 X_{AC}}{X_{AC} C_1}\right| \left|\frac{C_1 X_{BC}}{X_{BC} B_1}\right| = 1. \end{align}Using again Menelaus this equation implies that \(X_{AB}\), \(X_{BC}\) and \(X_{AC}\) are collinear.
Non-Desarguesian planes
In the first section we did not specify further what we meant with "the projective plane" and just took it to be the usual extended Euclidean plane. Indeed, Desargues' theorem holds in this plane, but there are other planes where the theorem does not hold. Such planes are possible because the purely two-dimensional proof of the theorem requires the use of the axioms of congruence. In a plane where some of these axioms are violated, Desargues' theorem does not hold. Such planes are called Non-Desarguesian planes.
In his book "The Foundations of Geometry" Hilbert constructs a Non-Desarguesian plane as follows. Fix an ellipse in the Euclidean plane with foci on the \(x\)-axis and major and minor axes of lengths \(1\) and \(2\) respectively. Fix also the point \(F = (3, 0)\) on the \(x\)-axis.
The points of the Non-Desarguesian plane are simply the points \((x, y)\) of the Euclidean plane. A straight line in the Euclidean plane that does not intersect the ellipse is also a straight line in the new plane. For a straight line that intersects the ellipse, we remove the segment between the intersection points and replace it by the circular arc through the intersection points and the third point \(F\). This is shown in the following graphic.
Hilbert shows that the circle through the intersection points \(Q_1\), \(Q_2\) and the third point \(F\) does not meet the ellipse in any other points, so this construction is well-defined (in fact \(F\) was chosen in order for this to work).
In the new geometry two distinct points still uniquely determine a straight line; moreover one can choose any two distinct points on a given line and those two points uniquely determine it. The parallel axiom also still holds.
But Desargues' theorem does not hold. To see this, consider three lines that intersect in the origin. Choose a point on each line and form the triangle \(\Delta_1 = A_1 B_1 C_1\). Draw another triangle \(\Delta_2\) with sides parallel to the ones of \(\Delta_1\). Each pair of parallel lines has an intersection point on the line at infinity. However, the new straight lines are deformed inside the ellipse and no longer meet in a single point:
It turns out that in order to prove Desargues' theorem in the plane, one has to assume the axioms of congruence.
References
- David Hilbert: The Foundations of Geometry
- The graphics were made with JSXGraph.