定理 1　三角形的内、重、界三心共线且重心在中间 ,重界距离等于重内距离的 2倍 .证明 :设△ABC的内心为I,重心为G ,界心为J ,M为BC的中点 ,连结AJ、MI、IJ,AM ,IJ与AM交于G′.由 [1 ]知AJ∥IM ,由[2 ]知 ,AJ=2IM ,从而AG′ =2G′M .可见G′就是重心G .进而知三心共线 ,且JG = Theorem 1 The triangle’s inner, center, and inner circles have three lines in common and the center of gravity is in the middle. The distance between the centers is equal to twice the distance in the inner circle. Proof: Let △ABC have an inner center of gravity I, a center of gravity is G, a center of gravity is J, and M is The midpoint of BC, which links AJ, MI, IJ, AM, IJ and AM, intersects G’. Known by [1] is AJ∥IM, known by [2], AJ=2IM, and thus AG’ = 2G’M. G’ is the center of gravity G. Then the three hearts are collinear, and JG =