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
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![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](/uploads/image/z/2449570-58-0.jpg?t=%E8%AE%BEO%E4%B8%BA%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%A4%96%E5%BF%83R%E4%B8%BA%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%A4%96%E6%8E%A5%E5%9C%86%E5%8D%8A%E5%BE%84%2COA%2COB%2COC%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E5%88%86%E5%88%AB%E4%BA%A4BC%2CCA%2CAB%E4%BA%8ED%2CE%2CF.%E6%B1%82%E8%AF%81%3A1%5CAD%2B1%5CBE%2B1%5CCF%3D2%5Cr)
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
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
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