题目内容
如图四棱锥P-ABCD中,ABCE为菱形,E、G、F分别是线段AD、CE、PB的中点.求证:FG∥平面PDC.
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证明:连接BD与CE交于点0,∵E为AD的中点,ABCE为菱形,AE=BC=DE,
∴
=
=1,得到O为线段CE的中点,故O与点G重合.
∵
=
=1,∴G为BD的中点,又F为PB的中点,
∴FG∥PD,又∵FG?平面PDC,PD?平面PDC.
∴FG∥平面PDC.
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∴
CO |
OD |
BC |
DE |
∵
BG |
GD |
BC |
ED |
∴FG∥PD,又∵FG?平面PDC,PD?平面PDC.
∴FG∥平面PDC.
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