设O为锐角三角形ABC的外心R为三角形ABC的外接圆半径,OA,OB,OC的延长线分别交BC,CA,AB于D,E,F.求证:1\AD+1\BE+1\CF=2\r

设O为锐角三角形ABC的外心R为三角形ABC的外接圆半径,OA,OB,OC的延长线分别交BC,CA,AB于D,E,F.
求证:1\AD+1\BE+1\CF=2\r

r/AD=S△ACO/S△ACD=S△ABO/ABD
=(S△ACO+S△ABO)/S△ABC
同理:
r/BE=(S△BAO+S△BCO)/S△ABC
r/CF=(S△CAO+S△CBO)/S△ABC
故:r/AD + r/BE + r/CF =2S△ABC/S△ABC=2
1\AD+1\BE+1\CF=2\r

很复杂
http://www.mrieka.com/qz/ea1856744.html