Since it is a result in the projective plane, it has a dual, Brianchon's Theorem, which states that the diagonals of a hexagon circumscribed about a conic concur.
It is sufficient to prove the result for a hexagon inscribed in a circle, for affine transformations map this circle to any ellipse while preserving collinearity and concurrence in the projective plane, and projective transformations can map an ellipse to any conic while similarly preserving collinearity and concurrence in the projective sense. Thus we will prove the theorem for a cyclic hexagon, using directed angles mod .
Lemma. Let be two circles which intersect at , let be a chord of , and let be the second intersections of lines with . Then and are parallel.
Proof. Since are two sets of concyclic points and and are two sets of collinear points,
Because alternate interior angles and are congruent, ∎
Theorem. Let be a cyclic hexagon, and let , , . Then are collinear.
Proof. Let be the circumcircle of , and let be the circumcircle of . Let be the second intersection of with , and let be the second intersection of with . By lemma, , and , and , see figure. It follows that triangles and are homothetic with a center , because . Therefore the line passes through the center of homothety, i.e. . ∎
In our proof, we never assumed anything about configuration. Thus the hexagon need not even be convex for the theorem to hold. In fact, many useful applications of the theorem occur with degenerate hexagons, i.e., hexagons in which not all of the points are distinct. In the case that two points are the same, we consider the line through them to be the tangent to the conic through that point. For instance, when we let a triangle be a "hexagon" , Pascal's Theorem tells us that if are the tangents to the circumcircle of that pass through , respectively, then , , are collinear; the line they determine is called the Lemoine Axis. In fact, Pascal's Theorem tells us that can be the tangent lines to any conic circumscribed about triangle and the result still holds.