如图,正方形ABCD所在平面与直角三角形ABE所在的平面互相垂直,AE⊥AB...
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发布时间：2024-10-24 08:03
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时间：2024-10-24 08:18
（Ⅰ）证明：取EC中点F，连接MF，BF．
∵MF为△CDE的中位线，
∴MF∥CD，MF＝12CD；
又∵NB∥CD，NB＝12CD，
∴NB∥MF，NB=MF
∴四边形NBFM为平行四边形，
∴MN∥BF，又∵BF?平面BEC，MN?平面BEC，
∴MN∥平面BEC；
（Ⅱ）∵MN∥平面BEC，
∴VE?BMC＝VM?BEC＝VN?BEC＝VC?BEN＝13S△BEN?CB＝13×12×2＝13
∵AB⊥AD，AB⊥AE，
∴AB⊥平面EAD，
∴AB⊥AM，
则MB＝MA2+AB2＝(12DE)2+AB2＝(52)2+22＝212
∵CD∥AB，
∴CD⊥平面EAD，故CD⊥DM，
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