如图,在△ABC中,BD=2DC,CE=2EA,AD与BE交于点F,且△ABC的面积为42,则△AEF的面积是多少

2024-12-03 09:08:31
推荐回答(1个)
回答1:

∵在△ABC中,BD=2DC,
∴CD=

1
3
BC,
∴S△ACD=
1
3
S△ABC=
1
3
×42=14,
∵CE=2EA,
∴AE=
1
3
AC,
S△ABE
S△ABC
=
AE
AC
=
1
3
S△AEF
S△ABE
=
EF
BE
=
1
7

S△AEF
SABC
=
1
21

∵S△ABC=42,
∴S△AEF=
42
21
=2.