Chongqing ae recruitment

∵ quadrilateral ABCD is a parallelogram,

∴AD∥BC,AD=BC，

∴∠EAD=∠AEB，

∫AE divided equally ∠BAD,

∴∠BAE=∠DAE，

∴∠BAE=∠BEA，

∴AB=BE，

AB = AE，

∴△ABE is an equilateral triangle; 2 correct;

∴∠ABE=∠EAD=60，

AB = AE，BC=AD，

∴△abc≌△aed(sas)； 1 correct;

∫△FCD and△ △ABC have the same base (AB=CD) and the same height (the distance between AB and CD is equal).

∴S△FCD=S△ABC，

∫△AEC and△ △DEC have the same base and the same height.

∴S△AEC=S△DEC，

∴s△abe=s△cef； ⑤ Correct.

∫AD and AF are not necessarily equal,

∴ ③ Not necessarily correct;

∫BE is not necessarily equal to ce,

∴ ④ Not necessarily correct.

So choose C.