解答:解:过点O作直线OM⊥AB于点M,交CD于点N.∵AB∥CD,∴ON⊥CD,∵AO是∠BAC角平分线,∴OM=OE=2,∵CO是∠ACD的角平分线,∴ON=OE=2,∴MN=2+2=4,即AB与CD之间的距离为4.故选D.