如图,等边△ABC的边长为2,F为AB中点,延长BC至D,使CD=BC,连接FD交AC于E,则四边形BCEF的面积为_.

问题描述:

如图,等边△ABC的边长为2,F为AB中点,延长BC至D,使CD=BC,连接FD交AC于E,则四边形BCEF的面积为______.

∵DEF是△ABC的梅氏线,
∴由梅涅劳斯定理得,

AF
FB
BD
CD
CE
EA
=1,
1
1
4
2
CE
EA
=1,则
CE
EA
=
1
2

连FC,S△BCF=
1
2
S△ABC,S△CEF=
1
6
S△ABC
于是SBCEF=S△BCF+S△CEF
=
2
3
S△ABC
=
2
3
×
1
2
×2×2sin60°
=
4
3
×
3
2
=
2
3
3

故答案为
2
3
3