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

Theorem

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.

Proof

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.

What seems like a surprising fact (Desargue's Theorem) becomes intuitively obvious once you don't restrict the points to a plane. The three points a, b and c determine a plane and similarly for a', b' and c'. The line ab and the line a'b' are confined to their respective planes and thus their intersection must lie on both of these planes. Similarly for the intersection of lines bc and b'c', and that of ac and a'c'. So we now have the three points of intersection lying simultaneously in 2 intersecting planes. Since 2 planes intersect in a line, the intersections are collinear. Q.E.D.

Exercise for the student: Where in the above did I use the concurrency of aa', bb', and cc'?

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