(1)连结AC交BD于O,连接EO 因为平行四边形ABCD, 所以O为BD中点,E为CC1中点 所以OE为△AC1C中位线, 所以OE∥AC1-----------3 OE⊂面BDE AC1⊄面BDE AC1∥面BDE------------6 (2)因正四棱柱ABCD-A1B1C1D1 所以BD⊥A1A,又因BD⊥AC A1A∩AC=AA1A⊂面A1AC C1 AC⊂面A1ACC1 所以BD⊥面A1ACC1----------------9 A1E⊂面A1ACC1 所以BD⊥A1E, A1E与BD所成角为900----------12 |