题目内容

已知四棱锥P-ABCD中,底面ABCD为菱形,∠ABC=60°,PA⊥平面ABCD,E为BC中点,求证:AE⊥PD.
证:设
AB
=
a
AD
=
b
AP
=
c

∵PA⊥平面ABCD,
a
c
=0,
b
c
=0,
∵∠ABC=60°,四边形ABCD为菱形,
a
b
=|
a
|•|
b
|•cos∠BAD=|
b
|2•cos120°
=-
1
2
|
b
|2
AE
=
AB
+
BE
=
a
+
1
2
b

PD
=
PA
+
AB
+
BC
+
CD
=-
c
+
a
+
b
-
a
=
b
-
c

AE
PD
=(
a
+
1
2
b
)•(
b
-
c

=
a
b
+
1
2
|
b
|2-
a
c
-
1
2
b
c

=-
1
2
|
b
|2+
1
2
|
b
|2=0,
AE
PD

∴AE⊥PD.
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