A Triangle Formed by the Centers of Three Nine-Point Circles

AB	≈	7.25	s(AB)	≈	1.87	A''B''	≈	3.63	s(A''B'')	≈	1.87
AC	≈	8.19	s(AC)	≈	0.22	A''C''	≈	4.10	s(A''C'')	≈	0.22
BC	≈	6.51	s(BC)	≈	-1.01	B''C''	≈	3.25	s(B''C'')	≈	-1.01

Let ABC be a triangle and let A', B', and C' be the midpoints of BC, AC, and AB, respectively. Let A'', B'', and C'' be the centers of the nine-point circles of the triangles AB'C', BC'A', and CA'B', respectively. Then A''B''C'' is homothetic with ABC in the ratio 1:2.

In the figure s(XY) is the slope of XY.