已知O是三角形ABC的外心,角BAC=45°,延长BC至D,使CD=1/2BC,AD平行OC,求∠ABC

Question

已知O是三角形ABC的外心,角BAC=45°,延长BC至D,使CD=1/2BC,AD平行OC,求∠ABC

Answer 1

连接OA、OB,延长BO交AD于E；
已知,O是△ABC的外心,可得：OA = OB = OC ,
所以,∠OAB = ∠OBA ,∠OAC = ∠OCA ,∠OBC = ∠OCB ；
因为,∠OAB+∠OBA+∠OAC+∠OCA = 2(∠OAB+∠OAC) = 2∠BAC = 90° ,
所以,∠OBC = ∠OCB = [180°-(∠OAB+∠OBA+∠OAC+∠OCA)] = 45° ,
可得：∠BOC = 180°-∠OBC-∠OCB = 90° ；
因为,AD∥OC,CD = ½BC ,即 CD/BC = 1/2 ,
所以,∠CAD = ∠OCA = ∠OAC ,∠BED = ∠BOC = 90° ,OE/OB = CD/BC = 1/2 ；
在△OAE中,∠OEA = 180°-∠BED = 90° ,OE = ½OB = ½OA ,可得：∠OAE = 30° ,
∠OAC = ∠CAD = ½∠OAE = 15° ,
∠OBA = ∠OAB = ∠BAC-∠OAC = 30° ,
∠ABC = ∠OBA+∠OBC = 75° .