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如图所示,在△ABC中,AD、BE相交于点O,BD:CD=3:2,AE:CE=2:1,若S△COD=2,求:(1)S△BOC:S△AOC:S△AOB的值.(2)求S△ABC.

题目详情
如图所示,在△ABC中,AD、BE相交于点O,BD:CD=3:2,AE:CE=2:1,若S△COD=2,求:

(1)S△BOC:S△AOC:S△AOB的值.
(2)求S△ABC
▼优质解答
答案和解析
(1)∵BD:CD=3:2,S△COD=2,
∴S△BOD=
2
2
×3=3,
S△BOC=2+3=5,
∵AE:CE=2:1,
AE
AC
=
2
2+1
=
2
3

过点E作EF∥BC交AD于F,则△AEF∽△ACD,
EF
CD
=
AE
AC
=
2
3

∵BD:CD=3:2,
EF
BD
=
2
3
÷
3
2
=
9
4

∵EF∥BC,
OF
OD
=
EF
BD
=
9
4

∵EF∥BC,
AF
DF
=
AE
EC
=2,
∴AF=2DF,
设OF=4k,OD=9k,
AF=2(OF+OD)=2(4k+9k)=26k,
∴AO=26k+4k=30k,
AO
OD
=
30
9
=
10
3

∴S△AOC=2×
10
3
=
20
3

S△AOB=3×
10
3
=10,
∴S△BOC:S△AOC:S△AOB=5:
20
3
:10=3:4:6;
(2)S△ABC=S△BOC+S△AOC+S△AOB=5+
20
3
+10=
65
3