[alogo] Hexagonal arc-polygon (2)

Continuation from SuccessiveArcsHex.html . Consider a circumscriptible hexagon p = ABCDEF. Start with an arbitrary point G on side EF and draw arcs, centered at the vertices and end-points on adjacent sides, so that the start-point of the next is the end-point of the previous. The construction results in an arc-polygon a1 = GJKNOR, closing back to the start point G (arc-polygon-1). In the same vein, one constructs the arc-polygon a2 = HILMPQ, of the contact points of the circumscriptible hexagon with its incircle (arc-polygon-2). In the above reference, it is shown that segments {HG, JI, LK, ... } are equal and a2 has its vertices on a circle concentric to the incircle of p. Here we investigate the triangle t1, formed by the diagonals of a1. It is shown that this triangle has its vertices on the diagonals of p. The same is true for the triangle t2, formed by the diagonals of a2. In addition the two triangles are anti-homothetic, the center of homothety being the intersection-point (Brianchon point) of the diagonals of p.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

Start with lines PI, MH and FC. They meet at a point T, according to Brianchon's theorem. Analogous arguments show that triangle t2 = STU, formed by the diagonals of a2, has its vertices on the diagonals of p. From the equality of segments (MN = HG), follows that NG, a diagonal of a1 is parallel to MH, diagonal of a2. Analogous remarks hold for the other diagonals too. Then lines NG, OJ, being parallel to MT, PT respectively, at equal distances, meet on the line CT, on which meet MT and PT (look at HomothetyProperty.html ). This implies all the statements formulated above.

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