题目内容

已知△ABC,点H,O为△ABC所在平面内的点,且
AH
•
AB
=
AH
•
AC
,
BH
•
BA
=
BH
•
BC
,
OA
+
OB
+
OC
=
OH
,则点O为△ABC的(  )
A.内心B.外心C.重心D.垂心
∵
AH
•
AB
=
AH
•
AC
,∴(
AB
-
AC
)•
AH
=0
即
CB
•
AH
=0
又
OA
+
OB
+
OC
=
OH
,∴
OB
+
OC
=
OH
-
OA
,即
AH
=
OB
+
OC
,
∴
CB
•(
OB
+
OC
)=0,即(
OB
-
OC
)•(
OB
+
OC
)=0,
∴|
OB
|2
=|
OC
|2
,∴OB=OB
同理OA=OC,
∴O是△ABC的垂心.
故选B.
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