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1) The quasi-Euler line of a quadrilateral: Given a quadrilateral ABCD, denote by Ga, Oa and Ha the centroid, circumcentre and orthocentre respectively of triangle BCD, and similarly, Gb, Ob, Hb for triangle ACD, Gc, Oc, Hc for triangle ABD, and Gd, Od, Hd for triangle ABC. Let G = GaGc ∩ GbGd, O = OaOc ∩ ObOd, H = HaHc ∩ HbHd. Then the quasi-orthocentre H, the lamina centroid G, and the quasi-circumcentre O are collinear. Furthermore, HG : GO = 2 : 1.
Reference: Myakishev, A. (2006). On Two Remarkable Lines Related to a Quadrilateral, Forum Geom., 6, 289–295.
Instructions: Drag any of the red vertices A, B. C or D below, and click on the buttons to show/hide.
The quasi-Euler line of a quadrilateral
2) The quasi-Euler line of a hexagon: Subdivide the hexagon ABCDEF into the six quadrilaterals ABCD, BCDE, CDEF, DEFA, EFAB, and FABC, and determine the quasi-circumcentre O, the lamina centroid G and quasi-orthocentre H of each quadrilateral. Then the main diagonals of each of the 3 hexagons formed by these centres are concurrent respectively in O, G and H. More-over, O, G and H are collinear and HG : GO = 2 : 1. (Label the respective hexagons as PQRSTU, PGQGRGSGTGUG and PHQHRHSHTHUH).
Instructions: Drag any of the red vertices A, B. C, D, E, or F below, and click on the buttons to show/hide.
The quasi-Euler line of a hexagon
Read my paper Quasi-circumcenters and a Generalization of the Quasi-Euler Line to a Hexagon (pdf, 45 KB, Forum Geometricorum, Vol 14(2014), pp. 233-236).
This interactive link includes more on the quasi-circumcentre of a quadrilateral and its analogue, the quasi-incentre.
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By Michael de Villiers. Created, 23 April 2014.