三角形ABC的外心为O,有一点H,向量OH=向量OA+向量OB+向量OC,求证H为重心注意是重心!

Question

三角形ABC的外心为O,有一点H,向量OH=向量OA+向量OB+向量OC,求证H为重心
注意是重心!

Answer 1

以下(a. b)表示a点乘b.
= = = = = = = = =
证明：设 向量OA = 向量x,
向量OB = 向量y,
向量OC = 向量z.
由已知,
OH =x+y+z.
且 |x| =|y| =|z|.
所以 AH =OH -OA =y+z,
BH =OH -OB =x+y,
CH =OH -OC =x+z.
又因为 BC =OC -OB =z-y,
CA =OC -OA =z-x,
AB =OB -OA =y-x,
所以 (AH. BC) =(y+z)(z-y) =z^2 -y^2 =0.
(BH. CA) =(x+y)(z-y) =z^2 -x^2 =0.
(CH. AB) =(x+y)(y-x) =y^2 -x^2 =0.
所以 AH 垂直于BC, BH垂直于CA, CH垂直于AB.
所以 H为三角形ABC的垂心.
= = = = = = = = =
外心的条件：
|OA| =|OB| =|OC|.
垂心的条件：
(AH. BC) =(BH. AC) =(CH. AB) =0.
取经常用到的向量OA,OB,OC为x,y,z, 把向量问题变成代数问题.