解:过E作EM⊥CD,反向延长交AB于点N,∵平行四边形ABCD中,OB=OD,E为OD的中点,∴BE=3DE,CD∥AB,∴△DEF∽△BEA,∴ ME EN = DE BE = 1 3 ,∴ ME MN = 1 4 ,∴ S△DEF S△DAF = 1 4 ,∴S△DEF:S△ADE=1:3.故选B.