离1031
幼苗
共回答了18个问题采纳率:88.9% 举报
(Ⅰ)设CD的中点为H,连结EH,
依题意得EH//PD,且EH=
![](https://img.yulucn.com/upload/a/94/a94aa0a25c69c902087222773266fd5d_thumb.jpg)
PD=1,因为PD⊥底面ABCD,所以EH⊥底面ABCD,故三棱锥E-ABD的高是EH,其体积为
因为
![](https://img.yulucn.com/upload/6/0f/60fd8b216e6289a0cbf8629339d61af5_thumb.jpg)
,所以三棱锥A-BDE的体积为
![](https://img.yulucn.com/upload/6/27/627d55a038c7e72b297d918a57096ce8_thumb.jpg)
.
(Ⅱ)证明:连结AC,AC交BD于O,连EO,∵底面 ABCD是正方形,∴点O是AC中点,在△PAC中,EO是中位线,∴PA∥EO,而EO
![](https://img.yulucn.com/upload/2/fb/2fbabf36ac39278a1092a6be36f253e8_thumb.jpg)
平面EDB,且PA
![](https://img.yulucn.com/upload/2/52/25213491a79e48816f53e77ee2f6e304_thumb.jpg)
平面EDB,∴PA∥平面EDB.
(Ⅲ) 证明:∵PD⊥底面ABCD且DC
![](https://img.yulucn.com/upload/2/fb/2fbabf36ac39278a1092a6be36f253e8_thumb.jpg)
底面ABCD,
∴PD⊥DC.
∵PD=DC可知△PDC是等腰直角三角形,而DE是斜边PC的中线,
∴DE⊥PC.①
同样由PD⊥底面ABCD,得PD⊥BC,
∵底面ABCD是正方形有DC⊥BC,
∴BC⊥平面PDC,而DE
![](https://img.yulucn.com/upload/2/fb/2fbabf36ac39278a1092a6be36f253e8_thumb.jpg)
平面PDC,
∴BC⊥DE.②
由①②得DE⊥平面PBC,而PB
![](https://img.yulucn.com/upload/2/fb/2fbabf36ac39278a1092a6be36f253e8_thumb.jpg)
面PBC,
∴DE⊥PB又EF⊥PB且DE∩EF=E,
∴PB⊥平面EFD.
略
1年前
7