Six Incircles in an Equilateral Triangle
From an interior point P of an equilateral triangle ABC draw lines perpendicular to the sides. Let A', B', and C' be the points on the sides opposite A, B and C. Inscribe circles in the six subtriangles. Let (XYZ) denote the diameter of the incircle in triangle XYZ.[more]
Then the sum of the diameters of the red circles equals the sum of the diameters of the blue circles:
(APC') + (BPA') + (CPB') = (C' PB) + (A' PC) + (B' PA)[less]
For more information see Six Incircles in an Equilateral Triangle.