Question

已知△ABC 为斜三角形,且O是△ABC所在平面上的一个定点,动点P满足向量OP=OA+入{(AB/|AB|＾2*sin

Answer 1

OP=OA+入{(AB/|AB|＾2*sin2B)+AC/(|AC|＾2*sin2C)},

OP-OA=入{(AB/|AB|＾2*sin2B)+AC/(|AC|＾2*sin2C)},

AP=入{(AB /|AB|＾2*sin2B)+AC /(|AC|＾2*sin2C)},

AP•BC=入{(AB•BC /|AB|＾2*sin2B)+AC•BC /(|AC|＾2*sin2C)},

AP•BC=入{|AB|•|BC|cos(180° -B) / (|AB|＾2*sin2B) +|AC|•|BC| cosC/(|AC|＾2*sin2C)},

AP•BC=入{-|AB|•|BC| cos B/ (|AB|＾2*2sinB cos B) +|AC|•|BC| cosC/(|AC|＾2*2sinC cosC)},

AP•BC=入{-|BC|/ (|AB|*2sinB ) +|BC|/(|AC|*2sinC )},

根据正弦定理得：|AB|/sinC=|AC|/ sinB,所以|AB|*sinB=|AC|*sinC

∴-|BC|/ (|AB|*2sinB ) +|BC|/(|AC|*2sinC )=0,

即AP•BC=0,

P点轨迹过三角形的垂心