9. 如图,在多面体ABCDEFABCDEFABCDEF中,四边形ABCDABCDABCD是一个矩形,EF//ACEF//ACEF//AC,AC=2EFAC=2EFAC=2EF,AB=AE=2AB=AE=2AB=AE=2,AD=4AD=4AD=4,∠BAE=120∘\angle BAE=120\circ∠BAE=120∘.(1)求证:AE//AE//AE//平面BFDBFDBFD;(2)若平面EAB⊥EAB\botEAB⊥平面ABCDABCDABCD,求平面EABEABEAB与平面FCDFCDFCD的夹角的余弦值.