Question

如圖,在梯形ABCD中,AB∥CD,AD=DC=CB=a,∠ABC=60°,平面ACFE⊥平面ABCD,四邊形ACFE是矩形,AE=a,點(diǎn)M在線段EF上.

(1)求證:BC⊥平面ACFE;

(2)當(dāng)EM為何值時(shí),AM∥平面BDF?證明你的結(jié)論.

Answer 1

解:(1)在梯形ABCD中,∵AB∥CD,AD=DC=CB=a,∠ABC=60°,∴四邊形ABCD是等腰梯形,且∠DCA=∠DAC=30°,∠DCB=120°.∴∠ACB=90°.∴AC⊥BC.又∵平面ACFE⊥面ABCD,交線為AC,∴BC⊥平面ACFE.

(2)當(dāng)EM=a時(shí),AM∥平面BDF.

在梯形ABCD中,設(shè)AC∩BD=N,連結(jié)FN,則CN∶NA=1∶2.∵EM=a,而EF=AC=a,

∴EM∶MF=1∶2.∴MF∥AN.∴四邊形ANFM是平行四邊形.∴AM∥NF.又∵NF平面BDF,AM平面BDF,∴AM∥平面BDF.