摘要: 如图,△ABC内接于⊙O,直径AD交BC于点E,延长AD至点F,使DF=2OD,连接FC并延长交过点A的切线于点G,且满足AG∥BC,连接OC,若cos∠BAC=,BC=6.
(1)求证:∠COD=∠BAC;
(2)求⊙O的半径OC;
(3)求证:CF
(1)求证:∠COD=∠BAC;
(2)求⊙O的半径OC;
(3)求证:CF
(1)求证:∠COD=∠BAC;
(2)求⊙O的半径OC;
(3)求证:CF是⊙O的切线.
【解答】解:(1)∵AG是⊙O的切线,AD是⊙O的直径,
∴∠GAF=90°,∵AG∥BC,∴AE⊥BC,∴CE=BE,∴∠BAC=2∠EAC,
∵∠COE=2∠CAE,∴∠COD=∠BAC;
(2)∵∠COD=∠BAC,∴cos∠BAC=cos∠COE==,∴设OE=x,OC=3x,
∵BC=6,∴CE=3,∵CE⊥AD,
∴OE2+CE2=OC2,∴x2+32=9x2,
∴x=(负值舍去),
∴OC=3x=,
∴⊙O的半径OC为;
- ∵DF=2OD,
∴OF=3OD=3OC,
∴,
∵∠COE=∠FOC,
∴△COE∽△FOE,
∴∠OCF=∠DEC=90°
∴CF是⊙O的切线.