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kaisibaoer
丨Lv 4
圆内接四边形四顶点组成的四个三角形重心构成的四边形与原四边形相似。 证明 记G1,G2,G3,G4分别是圆内接四边形A1A2A3A4中△A2A3A4,△A3A4A1,△A4A1A2,△A1A2A3的重心.E为A3A4的中点,联结A1E,A2E.则G1在A2E上,G2在A1E上,于是在△A1EA2中,G1G2=A1A2/3,G1G2∥A1A2.同理可得: G2G3=A2A3/3,G2G3∥A2A3;G3G4=A3A4/3,G3G4∥A3A4;G4G1=A4A1/3,G4G1∥A4a1.由此可得:∠G1G2G3=∠A1A2A3;∠G2G3G4=∠A2A3A4;∠G3G4G1=∠A3A4A1;∠D4D1D2=∠A4A1A2.因此四边形G1G2G3G4∽四边形A1A2A3A4,且相似比为1:3.