CD平行FE平行AB,CD=2,AB=4,EF将梯形ABCD的面积两等分,求FE的长

(CD + EF)h1 = 1/2*(CD+AB)h
(
h1：EF,CD构成的梯形的高,
h：原梯形的高
)
=>
(2+EF)h1 = 3h
=>
h1/h = 3/(2+EF)
过D做DG平行于BC交EF,AB于G,H
=>
DCBH为平行四边形
BH=CD=2
=>
AH=4-2=2
EG=EF-CD=EF-2
h1,h分别为三角形DEG,DAH的高
三角形DEG相似于DAH
=>
h1/h = EG/AH = (EF-2)/(4-2) = (EF-2)/2
代入上面的比例式得
=>
3/(2+EF) = (EF-2)/2
=>
EF=√10