题目内容
如图在平行六面体ABCD-A1B1C1D1中,E、F、G分别是A1D1、D1D、D1C1的中点.
求证:平面EFG∥平面AB1C.
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求证:平面EFG∥平面AB1C.
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证明:设
=a,
=b,
=c,则
=
+
=
(a+b),
=a+b=2
,
∴
∥
,
=
+
=
b-
c=
(b-c),
=
+
=b-c=2
,
∴
∥
.
又∵EG与EF相交,AC与B1C相交,
∴平面EFG∥平面AB1C.

AB |
AD |
AA1 |
EG |
ED1 |
D1G |
1 |
2 |
AC |
EG |
∴
EG |
AC |
EF |
ED1 |
D1F |
1 |
2 |
1 |
2 |
1 |
2 |
B1C |
B1C1 |
C1C |
EF |
∴
EF |
B1C |
又∵EG与EF相交,AC与B1C相交,
∴平面EFG∥平面AB1C.

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