O是三角形ABC外接圆的圆心连接AO交BC于D连接BO交AC于E连接CO交AB于F,R为半径,求证:1/AD+1/BE+1/CF=2/R
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![O是三角形ABC外接圆的圆心连接AO交BC于D连接BO交AC于E连接CO交AB于F,R为半径,求证:1/AD+1/BE+1/CF=2/R](/uploads/image/z/6870916-28-6.jpg?t=O%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E5%A4%96%E6%8E%A5%E5%9C%86%E7%9A%84%E5%9C%86%E5%BF%83%E8%BF%9E%E6%8E%A5AO%E4%BA%A4BC%E4%BA%8ED%E8%BF%9E%E6%8E%A5BO%E4%BA%A4AC%E4%BA%8EE%E8%BF%9E%E6%8E%A5CO%E4%BA%A4AB%E4%BA%8EF%2CR%E4%B8%BA%E5%8D%8A%E5%BE%84%2C%E6%B1%82%E8%AF%81%EF%BC%9A1%2FAD%2B1%2FBE%2B1%2FCF%EF%BC%9D2%2FR)
O是三角形ABC外接圆的圆心连接AO交BC于D连接BO交AC于E连接CO交AB于F,R为半径,求证:1/AD+1/BE+1/CF=2/R
O是三角形ABC外接圆的圆心连接AO交BC于D连接BO交AC于E连接CO交AB于F,R为半径,求证:1/AD+1/BE+1/CF=2/R
O是三角形ABC外接圆的圆心连接AO交BC于D连接BO交AC于E连接CO交AB于F,R为半径,求证:1/AD+1/BE+1/CF=2/R
设ha,hb,hc分别是△ABC边BC,CA,AB上的高,BC=a,CA=b,AB=c,S表示其面积.
∵1/AD=cos(B-C)/ha,1/BE=cos(C-A)/hb,1/CF=cos(A-B)/hc
S=(sinA*bc)/2=2R^2*4sinA*sinB*sinC
∴1/AD+1/BE+1/CF
=a*cos(B-C)/(2S)+b*cos(C-A)/(2S)+c*cos(A-B)/(2S)
=[a*cos(B-C)+b*cos(C-A)+c*cos(A-B)]/(2S)
=R*[sinA*con(B-C)+sinB*cos(C-A)+sinC*cos(A-B)]/S
=(R/S)*[sin(B+C)*con(B-C)+sin(C+A)*cos(C-A)+sin(A+B)*cos(A-B)]
=(R/S)*[sin(2A)+sin(2B)+sin(2C)
=(R/S)*[4sinA*sinB*sinC]
=(R/S)*[S/(2R^2)]=2/R.