# Maxwell Theorem via the Center of Gravity

### Michel Cabart

30 April, 2008

Below we offer a proof of Maxwell's theorem that is based on the notion of barycenter. Maxwell's theorem states the following fact:

Given ΔABC and a point G, the sides of ΔMNP are parallel to the cevians in ΔABC through G. Then the cevians in ΔMNP parallel to the sides of ΔABC are concurrent.

Below, the vector joining point A to B will be written in bold, so that, for example, **AB** = - **BA**.

In ΔABC with A', B', C' on sides opposite the vertices A, B, and C, the fact that the cevians AA', BB', CC' concur in point G is equivalent to either of the two conditions:

There is a triple of real numbers | |

(1) | aGA + bGB + cGC = 0 |

There is a triple of real numbers | |

(2) |
A' = Z(B, b; C, c) B' = Z(A, a; C, c) C' = Z(A, a; B, b), |

where Z(X, x; Y, y) denotes the *barycenter* of two *material points* X and Y with masses x at X and y at Y.

Let's suppose lines AA', BB' and CC' intersect.

**Step 1**: NP, PM, MN are parallel to GA, GB, GC means there exists x, y, z such that **NP** = x**GA**,**PM** = y**GB**,**MN** = z**GC**.**NP** + **PM** + **MN** = **0**,**GA** + y**GB** + z**GC** = **0**.**NP** = a**GA**,**PM** = b**GB**,**MN** = c**GC**.

**Step 2**: MM', NN', PP' are parallel to BC, AC, AB meaning there is **MM'** = m**BC**,**NN'** = n**AC**,**PP'** = p**AB**.

**MM'** = m(**GC** - **GB**) = (m/c)**MN** + (m/b)**MP**

so that

M' = Z(N, 1/c; P, 1/b) = Z(N, b; C, c)

by multiplying by bc. Similarly,

N' = Z(M, a; P, c) and

P' = Z(M, a; N, b).

This proves the theorem thanks to (2).

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