1701010520
设,且O为ΔABC的重心,则,且,记为S,那么,
1701010521
1701010522
1701010523
1701010524
1701010525
即:
1701010526
1701010527
1701010528
1701010529
1701010530
1701010531
1701010532
同理可得,。
1701010533
1701010534
1701010535
所以,则:
1701010536
1701010537
1701010538
1701010539
1701010540
3.过重心截取三角形两边长,形成的比例的倒数和为定值
1701010541
1701010542
1701010543
【问题】如图9-18所示,已知点O是ΔABC的重心,过O作直线与AB、AC两边分别交于M、N两点,且,则:
1701010544
1701010545
1701010546
1701010547
1701010548
1701010549
1701010550
1701010551
图9-18 三角形重心应用
1701010552
1701010553
【分析】
1701010554
1701010555
点O是ΔABC的重心,知:
1701010556
1701010557
1701010558
1701010559
1701010560
1701010561
得,从而有:
1701010562
1701010563
1701010564
1701010565
1701010566
又M、N、G三点共线(A不在直线AM上),于是存在λ和μ,使得:
1701010567
1701010568
1701010569
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