△ABC中,∠ABC和∠ACB的平分线BD,CE相交于点O,∴∠DAO=∠EAO,∠COD=∠OBC+∠OCB=(∠B+∠C)/2=(180°-∠A)/2=60°=∠A,∴A,D,O,E四点共圆,∴OE=OD.