如图,设O为△ABC内一点,连接AO、BO、CO,并延长交BC、CA、AB于点D、E、F,已知S△AOB:S△BOC:S△AOC=3:4:6.则ODAO•OEBO•OFCO等于(  ) A.235 B.435 C.635 D.835

问题描述:

如图,设O为△ABC内一点,连接AO、BO、CO,并延长交BC、CA、AB于点D、E、F,已知S△AOB:S△BOC:S△AOC=3:4:6.则

OD
AO
OE
BO
OF
CO
等于(  )
A.
2
35

B.
4
35

C.
6
35

D.
8
35

∵S△AOB:S△BOC:S△AOC=3:4:6,
∴S△AOB:S△ABC=3:13,S△BOC:S△ABC=4:13,S△AOC:S△ABC=6:13,

OF
CF
=
3
13
OD
AD
=
4
13
OE
BE
=
6
13

OF
CO
=
3
10
OD
AO
=
4
9
OE
BO
=
6
7

OD
AO
OE
BO
OF
CO
=
3
10
×
4
9
×
6
7
=
4
35

故选:B.