题目如图,平行四边形ABCDABCDABCD的对角线ACACAC,BDBDBD交于点OOO,AB⊥ACAB\bot ACAB⊥AC,AB=3AB=\sqrt{3}AB=3,∠AOB=60∘\angle AOB=60^{\circ}∠AOB=60∘,过点OOO作OE⊥ACOE\bot ACOE⊥AC,交ADADAD于点EEE,过点EEE作EF⊥BDEF\bot BDEF⊥BD,垂足为FFF,则OE+2EFOE+2EFOE+2EF的值为______.知识点:解分式方程章节:方程与不等式 / 分式方程