Question

高一数学必修四 平面向量的线性运算

O是三角形ABC内一点,满足向量OA+向量OB+向量OC=0,|向量OA|=|向量OB|=|向量OC|,求证△ABC是正三角形

Answer 1

let
|OA|=|OB|=|OC| = k

OA+OB+OC = 0
OA.OA = (OB+OC).(OB+OC)
k^2 = 2k^2 +2OB.OC
=> OB.OC = -k^2/2
similarly
OC = -(OA+OB)
OA.OB = -k^2/2
and
OC.OA = -k^2/2

AB = OB-OA
|AB|^2 = (OB-OA).(OB-OA)
= |OB|^2+ |OA|^2 - 2OB.OA
= 3k^2
|BC|^2 = (OC-OB).(OC-OB)
= |OB|^2+ |OC|^2 - 2OB.OC
= 3k^2
|CA|^2 = |OC|^2+|OA|^2 - 2OA.OC
= 2k^2 - 2OA.OC
= 3k^2
=>|AB|^2=|BC|^2=|CA|^2
=>|AB|=|BC|=|CA|
=> △ABC是正三角形