证明：连接OB,设DE交AB于F∵O是△ABC的外心,EF垂直平分AB∴O、D、E三点共线,∠AOF＝∠BOF＝∠AOB/2∵∠AOB为圆周角∠ACB的圆心角∴∠ACB＝∠AOB/2∴∠ACB＝∠AOF∵∠AOD＝180-∠AOF,∠DCE＝180-∠ACB∴∠AOD＝∠DCE...