The quasi-Euler line of a quadrilateral and a hexagon

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1) The quasi-Euler line of a quadrilateral: Given a quadrilateral ABCD, denote by G_{a}, O_{a} and H_{a} the centroid, circumcentre and orthocentre respectively of triangle BCD, and similarly, G_{b}, O_{b}, H_{b} for triangle ACD, G_{c}, O_{c}, H_{c} for triangle ABD, and G_{d}, O_{d}, H_{d} for triangle ABC. Let G = G_{a}G_{c} ∩ G_{b}G_{d}, O = O_{a}O_{c} ∩ O_{b}O_{d}, H = H_{a}H_{c} ∩ H_{b}H_{d}. 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 & a hexagon

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, P_{G}Q_{G}R_{G}S_{G}T_{G}U_{G} and P_{H}Q_{H}R_{H}S_{H}T_{H}U_{H}).

Instructions:
1) Use the 'Link to quasi-euler hexagon' button to navigate to the result formulated above.
2) Drag any of the red vertices A, B. C, D, E, or F below, and click on the buttons to show/hide.