Solution: If the intersection of AD and BC extends to F, DECF is a parallelogram.

∫EC∨AD，DE∨BC，

∴∠ADE=∠DEC=∠BCE,∠CBE=∠AED，

∴△CBE∽△DEA，

∫S△BEC = 1，S△ADE=3

∴BEAE= 13=33，

CEDF is a parallelogram,

∴△CDE≌△DCF，

∴S？ CEDF=2S△CDE

∫EC∨AD，

∴△BCE∽△BFA，

∴ bee +BE = 33+3, S △ BCE: S △ BFA = (33+3) 2, that is,1:(1+3+2s △ CDE) = 3 (3+3) 2.

Solution: s △ CDE = 3.

So choose C.