Michel 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 x > 0 if triangles are external and x < 0 if they are internal. (For a generic x, this is the contents of Kiepert's teorem.) Its barycentric coordinates are a/sin(A + x), b/sin(B + x), c/sin(C + x) so that the point is the affine sum
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 a[(cos(B - C) - cos(B + C + 2x)], etc. If the apex angle of the isosceles triangles is 90°, we get ordinary Vecten's points; for the apex angles of 120°, we get Napoleon's points.
Now let's consider two such Kiepert points with algebraic angles x and y such that x - y = 90°. The first Kiepert point, say K1 = K(x), has barycentric coordinates a[(cos(B - C) - cos(B + C + 2x)], etc. The second Kiepert point K2 = K(y) has barycentric coordinates a[(cos(B - C) - cos(B + C + 2y)], etc. or equivalently a[(cos(B - C) + cos(B + C + 2x)], etc.
The center of nine points circle, N, has barycentric coordinates [a·cos(B - C)], etc. It follows that
2·s3·N = s1·K1 + s2·K2,
where s1, s2, s3 being the sums of barycentric coordinates of K1, K2, and N. As 2·s3 = s1 + s2, N coincides with the barycenter of (K1, s1; K2, s2) and, hence, is on the line K1K2. (Put differently, in the determinant of barycentric coordinates of K1, K2, and N, the lines L1, L2, L3 satisfy L1 + L2 - 2L3 = 0 hence the determinant is zero and the lines are concurrent.)
