Question

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

Answer 1

∵∠BAC=45°

∴∠BOC=90°

设BO延长交AD于E

∵AD平行OC CD=1/2BC

∴OD=1/2DA ∠ADO=90° ∠OCA=∠DAC

∠AOD=60 °∠DOA=30°

∵OA=OB=OC

∴∠OBC=∠OCB=45°

∴∠OCA=∠OAC=1/2∠DOA=15°

∴∠OBA=∠OAB=∠BAC-∠OAC=45°-15°=30°

∴∠ABC=∠OBC+∠OBA=75°