4.如图,在四棱柱ABCD−A1B1C1D1ABCD-A_{1}B_{1}C_{1}D_{1}ABCD−A1B1C1D1中,侧棱A1A⊥A_{1}A\botA1A⊥平面ABCDABCDABCD,AB//DCAB//DCAB//DC,AB⊥ADAB\bot ADAB⊥AD,AD=CD=2AD=CD=2AD=CD=2,AA1=AB=4AA_{1}=AB=4AA1=AB=4,EEE为棱AA1AA_{1}AA1的中点.(1)证明:BC⊥C1EBC\bot C_{1}EBC⊥C1E.(2)设CM→=λCE→(0<λ<1)\overrightarrow{CM}=\lambda \overrightarrow{CE}(0<\lambda <1)CM=λCE(0<λ<1),若C1C_{1}C1到平面BB1MBB_{1}MBB1M的距离为255\frac{2\sqrt{5}}{5}525,求λ\lambdaλ.