梯形ABCD中,AD//EF//BC,EF将梯形ABCD两等分 AD=2,BC=4,求EF长

延长BA、CD交于M
显然△MAD∽△MEF∽△EBC
设S△MAD＝X,S梯形AEFD＝S梯形EBCF＝Y
则X/(X＋Y)＝(AD/EF)^2=4/EF^2
(X+Y)/(X+2Y)=(EF/BC)^2=EF^2/16
上述两式相乘得：X/(X+2Y)=1/4
解得X＝2Y/3
将X＝2Y/3代入X/(X＋Y)＝4/EF^2解得
EF＝√10