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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