在等腰梯形ABCD中,AB//CD,
∴AC=BD,
作DH⊥BC于H,设AH=x,则CD=AB-2x,
BD^2=DH^2+BH^2=AD^2-x^2+(AB-x)^2=AD^2+AB^2-2xAB,
在双曲线中,2c=AB,2a=BD-AD,离心率为c/a=AB/(BD-AD)=2.
在椭圆中,2c'=CD,2a'=AC+AD,e=c'/a'=CD/(AC+AD)=CD/(BD+AD)
=AB*CD/[2(BD-AD)(BD+AD)]
=AB*CD/[2(BD^2-AD^2)]
=CD/[2(AB-2x)]=1/2,
选A.