如图,已知平行四边形ABCD中,对角线AC、BD交于点O,E是DB延长线上一点,且△ACE是等边三角形

∵△ACE是等边三角形,∠EAC=∠AEC=∠ACE=60° AE=CE OE=EO AO=OC△AOE≌△COE所以OE平分∠AEC∴∠AED=30°∠AED=2∠EAD∠EAD=15°∴∠DAC=60-15=45°=∠CAB∴∠DAB=2*45=90°∵OE平分等边三角形∠AEC∴OE⊥ACAO=OC ∠AOD=∠COD OD=OD所以△AOD≌△CODAD=CDAB⊥AD∴平行四边形ABCD是正方形