The theorem and proof are taken from Dan Pedoe's Geometry.


Let A, B, and C, and also A', B', and C'--by grouping them this way you can visualize the theorem in terms of triangles--be position vectors of distinct points in a plane. Suppose that the lines AA', BB', and CC' are concurrent--they intersect at a single point V. Let X, Y, and Z be the intersections of BC and B'C', CA and C'A', and AB and A'B' respectively, if they exist (i.e. the sides of the triangle are not parallel). Then X, Y, and Z are collinear.


V can be written as rA + (1-r)A' = sB + (1-s)B' + tC + (1-t)C'. Then sB - tC = (1-t)C' - (1-s)B'. If X exists, then it can be written as pB + (1-p)C and as qB + (1-q)C'. Therefore sB - tC = (s-t)X, and similarly tC - rA = (t-r)Y and rA - sB = (r-s)Z.

Adding these three equations yields (s-t)X + (t-r)Y + (r-s)Z = 0. Not that the sum of the coefficients also equals 0. It can be shown that there are two possibilities in such a case---either X, Y, and Z are collinear or (s-t) = (t-r) = (r-s) = 0. But if r=s=t then all of the triangle lines are parallel. Since we assumed the intersections X, Y, Z existed, this cannot be the case, so X, Y, Z must be collinear.