AE:AB:BE=√6:√3:1,简要理由如下:
连结AF,设小圆半径为R,
∵AC是小圆的切线,
∴OB⊥AC,
∴AB=BC,
又∵FO=OC,
∴AF=2OB=2R,
∵OA⊥EF,
∴弧AE=弧AF,
∴AE=AF=2R,
由△OAB∽△FBA得OB/BA=AB/FA,
∴AB²=OB*AF=2R²,
∴AB=√2R,
由RT△OAB得OA=√3R,AG=2√3/3R,BG=√6/3R,
由RT△AEG得EG=2√6/3R,
∴BE=√6/3R,
∴AE:AB:BE=√6:√3:1
AE:AB:BE=√6:√3:1,简要理由如下:
连结AF,设小圆半径为R,
∵AC是小圆的切线,
∴OB⊥AC,
∴AB=BC,
又∵FO=OC,
∴AF=2OB=2R,
∵OA⊥EF,
∴弧AE=弧AF,
∴AE=AF=2R,
由△OAB∽△FBA得OB/BA=AB/FA,
∴AB²=OB*AF=2R²,
∴AB=√2R,
由RT△OAB得OA=√3R,AG=2√3/3R,BG=√6/3R,
由RT△AEG得EG=2√6/3R,
∴BE=√6/3R,
∴AE:AB:BE=√6:√3:1