题目内容
如图,平面
平面ABCD,ABCD为正方形,
是直角三角形,且
,E、F、G分别是线段PA,PD,CD的中点.
(1)求证:
∥面EFC;
(2)求异面直线EG与BD所成的角;



(1)求证:

(2)求异面直线EG与BD所成的角;

(1)证明见解析(2)

(1)证明:取AB中点H,连结GH,HE,∵E,F,G分别是线段PA、PD、CD的中点,∴GH∥AD∥EF,∴E,F,G,H四点共面,又H为AB中点,∴EH∥PB.又
面EFG,PB
面EFG,∴PB∥面EFG.………6分
(2)取BC的中点M,连结GM、AM、EM,则GM∥BD,
∴∠EGM(或其补角)就是异面直线EG与BD所成的角.
在Rt△MAE中,
,同理
,
又
,∴在MGE中,
,
故异面直线EG与BD所成的角为
.………………12分


(2)取BC的中点M,连结GM、AM、EM,则GM∥BD,
∴∠EGM(或其补角)就是异面直线EG与BD所成的角.
在Rt△MAE中,


又


故异面直线EG与BD所成的角为


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