Question

如图,在△ABC中,∠B=60°,角平分线AE、CF分别交BC、AB于E、F两点,AE、CF相交于点O.

Answer 1

证明：在AC边上取点G,使AG＝AF,连接OG∵∠B＝60∴∠BAC+∠ACB=180-∠B=120∵AE平分∠BAC,CF平分∠ACB∴∠CAE＝∠BAE＝∠BAC/2, ∠ACF＝∠BCF＝∠ACB/2∴∠AOF＝∠COE＝∠CAE+∠ACF＝（∠BAC+∠ACB）/2＝120/2＝60∴...