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

Question

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

Answer 1

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.