如图,在正方形ABCD中,点E是BC边上一点,且BE:EC=2:1,AE与BD交于点F,则△AFD与四边形DEFC的面积之比是_.

问题描述:

如图,在正方形ABCD中,点E是BC边上一点,且BE:EC=2:1,AE与BD交于点F,则△AFD与四边形DEFC的面积之比是______.

设CE=x,S△BEF=a,
∵CE=x,BE:CE=2:1,
∴BE=2x,AD=BC=CD=AD=3x;
∵BC∥AD∴∠EBF=∠ADF,
又∵∠BFE=∠DFA;
∴△EBF∽△ADF
∴S△BEF:S△ADF=(

BE
AD
)2=(
2x
3x
)
2
=
4
9
,那么S△ADF=
9
4
a.
∵S△BCD-S△BEF=S四边形EFDC=S正方形ABCD-S△ABE-S△ADF
9
2
x2-a=9x2-
1
2
×3x•2x-
9
4
a

化简可求出x2=
5
6
a

∴S△AFD:S四边形DEFC=
9
4
a
(
9
2
x2−a)
=
9
4
a
11
4
a
=9:11,故答案为9:11.