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已知O是△ABC的内一点,求证O是△ABC的重心的充要条件是OA+OB+OC=0向量解法
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已知O是△ABC的内一点,求证O是△ABC的重心的充要条件是OA+OB+OC=0
向量解法
向量解法
▼优质解答
答案和解析
必要性证明:设O为重心,E为BC中点.
OA=(2/3)EA==(2/3)(EB+BA)==(2/3)(CB/2+BA)=(CB+2BA)/3
同理,OB=(AC+2CB)/3.OC=(BA+2AC)/3.
CA+OB+OC=(3CB+3BA+3AC)/3=CC=0.
充分性证明:如图:OA={-x,-y}.OB={a-x.-y}.OC={b-x,c-y}.
OA+OB+OC={-x+a-x+b-x,-y-y+c-y}=0
-x+a-x+b-x=0.x=(a+b)/3.,-y-y+c-y=0,y=c/3.即O((a+b)/3,c/3)
请 564663878 朋友 自己验证.O((a+b)/3,c/3)正是⊿ABC的重心.
OA=(2/3)EA==(2/3)(EB+BA)==(2/3)(CB/2+BA)=(CB+2BA)/3
同理,OB=(AC+2CB)/3.OC=(BA+2AC)/3.
CA+OB+OC=(3CB+3BA+3AC)/3=CC=0.
充分性证明:如图:OA={-x,-y}.OB={a-x.-y}.OC={b-x,c-y}.
OA+OB+OC={-x+a-x+b-x,-y-y+c-y}=0
-x+a-x+b-x=0.x=(a+b)/3.,-y-y+c-y=0,y=c/3.即O((a+b)/3,c/3)
请 564663878 朋友 自己验证.O((a+b)/3,c/3)正是⊿ABC的重心.
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