Given that the center of gravity of △ ABC is g, GA = 3, GB = 4, GC = 5, calculate the area of △ ABC

Let ad, be, CF be the center line, it is easy to prove that Ag = 2Gd, that is, the center of gravity is a trisection point, extend GD to 0, so that DG = OD, connect B0, then og = 2Gd = Ag = 3 △ bd0 ≌ △ CDG, so B0 = CG = 5 in △ BGO, og = 3, BG = 4, B0 = 5, so △ BGO is a right triangle, s △ BGO = 1 / 2 * 3 * 4 = 6, so s △ BDG = 1 / 2 * s △ BGO = 3

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