Projective Proof of Maxwell's Theorem
Maxwell's theorem links the sides and cevians in one triangle to the cevians and side lines in another:
Given two triangles ABC and MNP. Assume that the cevians in ΔABC parallel to the sides of ΔMNP are concurrent. Then the cevians in ΔMNP parallel to the sides of ΔABC are also concurrent.
Michel Cabart came up with a generalization that shows that Maxwell's theorem is of projective nature. The main tool is the projective reformulation of Ceva's theorem.
By Michel Cabart
9 May, 2008
|(1)||R = (CBA'A0)·(ACB'B0)·(BAC'C0) = -1|
where (WXYZ) denotes the cross-ratio
In addition, let A1, B1, C1 be the intersection with L of the cevians AA', BB', and CC'.
(CBA'A0) = A(B0C0A1A0)
(ACB'B0) = B(C0A0B1B0),|
(BAC'C0) = C(A0B0C1C0)
Thus the condition (1) can be written (cross-ratios being unchanged):
|(2)||R = (A0B0C1C0)·(B0C0A1A0)·(C0A0B1B0) = -1|
Proof of Maxwell's theorem
Assume there are two triangles ABC and MNP and two triples of points on L such that A0, B0, C0 are intersections of L with sides of ΔABC and also with cevians of ΔMNP; A1, B1, C1 are intersections of L with sides of ΔMNP and also with cevians of ΔABC.
For ΔABC, we calculate
R0 = (A0B0C1C0)·(B0C0A1A0)·(C0A0B1B0).
For ΔMNP, we calculate
R1 = (A1B1C0C1).(B1C1A0A1).(A1B1C0C1)
We have the identity R0·R1 = 1 (as can be seen by developing cross-ratios). Hence
Note: a dynamic illustration is available on a separate page.
Copyright © 1996-2018 Alexander Bogomolny