Question

如图,三角形ABC与三角形DEC是等边三角形,AE,BD交于点O

Answer 1

证明：∵等边△ABC,等边△DCE∴AC＝BC,DC＝EC,∠BAC＝∠ABC＝∠ACB＝∠DCE＝60∵∠ACE＝∠DCE+∠ACD,∠BCD＝∠ACB+∠ACD∴∠ACE＝∠BCD∴△ACE≌△BCD （SAS）∴AE＝BD,∠CAE＝∠CBD∴∠AOD＝∠BAE+∠ABD＝∠BAC+∠CAE+∠ABD＝∠BAC+∠CBD+∠ABD＝∠BAC+∠ABC＝120°

180-120=60