题目已知在△ABC\triangle ABC△ABC中,AB=AC=4AB=AC=4AB=AC=4,∠BAC=120∘\angle BAC=120^{\circ}∠BAC=120∘,点DDD是CACACA延长线上任意一点,作DE⊥ABDE\bot ABDE⊥AB于点EEE,DF⊥BCDF\bot BCDF⊥BC于点FFF,连接EFEFEF,则EFEFEF的最小值为( )A.3\sqrt{3}3B.3−1\sqrt{3}-13−1C.433\frac{4}{3}\sqrt{3}343D.5−1\sqrt{5}-15−1知识点:直角三角形全等的判定章节:图形的性质 / 三角形