题目内容

平行四边形ABCD中AC交BD 于O,AC=5,BD=4,则(
AB
+
DC)
(
BC
+
AD
)
=(  )
A.41B.-41C.9D.-9

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如图所示,
DC
=
AB
=
OB
-
OA
BC
=
AD
=
OD
-
OA
=-
OB
-
OA

(
AB
+
DC)
(
BC
+
AD
)
=2(
OB
-
OA
)•2(-
OB
-
OA

=4(
OA
2
-
OB
2
)=4[(
5
2
)2-(
4
2
)2
]=9
故选C
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