Question

已知ABCD为矩形,SA⊥平面ABCD,过A作AE⊥SB于E,过E作EF⊥SC于F,如图所示,

(1)求证:AF⊥SC;

(2)若平面AEF∩SD=G,求证:AG⊥SD.

Answer 1

证明:(1)∵SA⊥平面ABCD,BC平面BCD,

∴SA⊥BC.又ABCD为矩形,

∴AB⊥BC.∴BC⊥平面SAB.

又∵AE平面SAB,∴BC⊥AE.

又∵由已知AE⊥SB,且SB∩BC=B,

∴AE⊥平面SBC.

∴AE⊥SC.

又∵AE∩EF=E,∴SC⊥平面AEF.

∴SC⊥AF.

(2)由(1)知SC⊥平面AEF,而AG平面AEF,

∴SC⊥AG.                                      ①

又∵SA⊥DC,AD⊥DC,SA∩AD=A,

∴DC⊥平面SAD.

∴DC⊥AG.                                    ②

由①②且DC∩SC=C,

∴AG⊥平面SDC.∴AG⊥SD.