最新中小学教案、试题、试卷
∴BC?平面EABF.又AF?平面EABF,∴BC?AF. (2)过M作MN?BC,垂足为N,连接FN,则MN∥AB. 又CM?
∴EF∥MN且EF?MN,∴四边形EFNM为平行四边形,∴EM∥FN. 又FN?平面FBC,EM?平面FBC,∴EM∥平面FBC. (3)由(1)可知,AF?BC.
在四边形ABFE中,AB?4,AE?2,EF?1,?BAE??AEF?90?, ∴tan?EBA?tan?FAE?11AC,∴MN?AB.又EF∥AB且EF?AB, 441,则?EBA??FAE. 2设AFIBE?P,∵?PAE??PAB?90?,
故?PBA??PAB?90?,则?APB?90?,即EB?AF. 又∵EBIBC?B,∴AF?平面EBC.