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

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.



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See problem 72 in N. Altshiller-Court, College Geometry, 2nd ed., Mineola, NY: Dover, 2007 p. 120.
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