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______.
答
∵DEF是△ABC的梅氏线,
∴由梅涅劳斯定理得,
•AF FB
•BD CD
=1,CE EA
即
•1 1
•4 2
=1,则CE EA
=CE EA
,1 2
连FC,S△BCF=
S△ABC,S△CEF=1 2
S△ABC,1 6
于是SBCEF=S△BCF+S△CEF
=
S△ABC2 3
=
×2 3
×2×2sin60°1 2
=
×4 3
=
3
2
.2
3
3
故答案为
.2
3
3