# Cabart's Collinearity

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Copyright © 1996-2018 Alexander BogomolnyMichel Cabart came up with a generalization of a result concerning Vecten's points and the 9-point center: the three are collinear. If Vecten triangles, i.e., asosceles right triangles, are replaced by two triples of isosceles triangles with supplementary apex (or complementary base) angles (but different orientations) then again the two points are collinear with the 9-point center.

The proof makes use of barycentric coordinates.

First it is easy to see by usual areas proof that a Vecten point exists when building 3 isosceles triangles with same base angles x on sides of triangle ABC, with

K(x) = A·a/sin(A + x) + B·b/sin(B + x) + C·c/sin(C + x).

Equivalently (eliminating denominators and using additive trigonometric formulas) the barycentric coordinates are

Now let's consider two such *Kiepert points* with algebraic angles x and y such that _{1} = K(x),_{2} = K(y)

The center of nine points circle, N, has barycentric coordinates

2·s_{3}·N = s_{1}·K_{1} + s_{2}·K_{2},

where s_{1}, s_{2}, s_{3} being the sums of barycentric coordinates of K_{1}, K_{2}, and N. As _{3} = s_{1} + s_{2},_{1}, s_{1}; K_{2}, s_{2})_{1}K_{2}. (Put differently, in the determinant of barycentric coordinates of K_{1}, K_{2}, and N, the lines L_{1}, L_{2}, L_{3} satisfy _{1} + L_{2} - 2L_{3} = 0

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Copyright © 1996-2018 Alexander Bogomolny67405106