已知正方体\(ABCD-{A}_{1}{B}_{1}{C}_{1}{D}_{1}\)的棱\({C}_{1}{D}_{1}\)上存在一点\(E\)（不与端点重合

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已知正方体\(ABCD-{A}_{1}{B}_{1}{C}_{1}{D}_{1}\)的棱\({C}_{1}{D}_{1}\)上存在一点\(E\)（不与端点重合），使得\(B{D}_{1}\)||平面\({B}_{1}CE\)，则（）

A.\(B{D}_{1}\)与\(CE\)不平行

B.\(A{C}_{1}\bot B{D}_{1}\)

C.\({D}_{1}E=2E{C}_{1}\)

D.\({D}_{1}E=E{C}_{1}\)

参考答案：AD