Desargues's theorem

Remarks

The two triangles are then said to be in perspective from a point and in perspective from a line. For two triangles in different planes of a projective space of dimension $3$, the proof is short: the three intersection points lie on both planes, hence on their common line. The plane case follows by viewing the plane figure as a projection of a figure in space, or by a computation with homogeneous coordinates. The converse direction is the dual statement, so it follows by the principle of duality.

In the Euclidean plane, if two triangles are homothetic, as $A' = 2A$, $B' = 2B$, $C' = 2C$ with centre the origin, the corresponding sides are parallel, so they meet on the line at infinity.

The theorem does not follow from the axioms of a projective plane alone. It fails in the projective completion of the Moulton plane and in other non-Desarguesian planes. It holds in a projective plane exactly when the plane is isomorphic to $\mathbb{P}^2(D)$ for some division ring $D$.

Going deeper

For a student

Desargues's theorem is the dividing line between planes that come from algebra and planes that do not. A projective plane in which it holds can be coordinatized by a division ring. It holds automatically in every projective space of dimension at least $3$, so the Veblen–Young theorem needs no extra axiom there. Pappus's hexagon theorem implies it, by Hessenberg's theorem, and Pappus holds exactly when the coordinate division ring is commutative. The little Desargues theorem, where the centre lies on the axis, characterizes Moufang planes.