详细问题描述及疑问:期待您的答案,感谢你,我会记得你对我的好的 !
【1】∵∠位大便病团CAB=60°∴∠ABC+∠BCA=180°-∠CAB=120°∵BE是∠ABC的平分线∴∠EBC=∠ABC同理∠BCD=∠BCA∵BE与CD交于F∴AF是∠CAB的平分线∠FBC=∠EBC∠FCB=∠BCD∴∠BFD=∠FBC+∠FCB=∠EBC+∠BCD=(∠ABC+∠BCA)=120°/2=60°=∠CA331215问答B∴ADFE四点共圆∴DF=EF【2】∵CD⊥AB于D,BD=CD∴∠ABC=45°∵BE⊥AC于E,BE与CD交于F∴AF⊥BC∴∠BAF=45°∴∠AFD=45°∵∠G=∠BAF∴∠D=∠AFD∵G在CD延长线上∴∠GAD=45°=∠BAF∴AGD=AFD∴GD=FD=DF