Napoleon's theorem

Drag the three vertices. An equilateral triangle stands outward on each side; the gold points are their centres N_a, N_b and N_c, joined into the gold triangle. However you drag, the longest over the shortest in the corner stays at 1.000000 and its centre's distance from the triangle's centroid G stays at 0: the gold triangle is always equilateral and centred at G. Choose outward and inward, and three more equilateral triangles stand inward, over the triangle itself; their centres make the blue triangle, also equilateral and also centred at G. The corner measures both areas, taking the triangle's as 1: gold minus blue is exactly 1. The closer the triangle gets to equilateral, the smaller the blue one becomes; when it is exactly equilateral, the blue one is a single point.