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设△ABC的外心为O,重心为G,取点H,使OH=OA+OB+OC.求证:(Ⅰ)点H为△ABC的垂心;(Ⅱ)△ABC的外心O、重心G、垂心H在同一条直线上.

题目详情
设△ABC的外心为O,重心为G,取点H,使
OH
OA
+
OB
+
OC
.求证:
(Ⅰ)点H为△ABC的垂心;
(Ⅱ)△ABC的外心O、重心G、垂心H在同一条直线上.
▼优质解答
答案和解析
证明:(Ⅰ)∵O为△ABC的外心,∴|
OA
|=|
OB
|=|
OC
|,
OH
OA
+
OB
+
OC
,∴
AH
OH
OA
OB
+
OC

AH
BC
=(
OB
+
OC
)•(
OC
OB
)=|
OC
|2−|
OB
|2=0
AH
BC
,即AH⊥BC,
同理BH⊥AC,CH⊥AB,
∴H为△ABC的垂心;
(Ⅱ)延长AG交BC于D,则D为BC中点,∴
AD
1
2
(
AB
+
AC
),
∵G为△ABC之重心,∴
AG
2
3
AD
1
3
(
AB
+
AC
)=
1
3
(
OB
+
OC
−2
OA
)
OG
OA
+
AG
OA
+
1
3
(
OB
+
OC
−2
OA
)=
1
3
(
OA
+
OB
+
OC
),
OH
=3
OG
,∴
OH
OG

∴O,G,H三点共线.