Question

如图,在梯形ABCD中,AB∥CD,AD=DC=CB=a,∠ABC=60°,平面ACFE⊥平面ABCD,四边形ACFE是矩形,AE=a,点M在线段EF上.

(1)求证:BC⊥平面ACFE;

(2)当EM为何值时,AM∥平面BDF?证明你的结论.

Answer 1

解:(1)在梯形ABCD中,∵AB∥CD,AD=DC=CB=a,∠ABC=60°,∴四边形ABCD是等腰梯形,且∠DCA=∠DAC=30°,∠DCB=120°.∴∠ACB=90°.∴AC⊥BC.又∵平面ACFE⊥面ABCD,交线为AC,∴BC⊥平面ACFE.

(2)当EM=a时,AM∥平面BDF.

在梯形ABCD中,设AC∩BD=N,连结FN,则CN∶NA=1∶2.∵EM=a,而EF=AC=a,

∴EM∶MF=1∶2.∴MF∥AN.∴四边形ANFM是平行四边形.∴AM∥NF.又∵NF平面BDF,AM平面BDF,∴AM∥平面BDF.