Proof: Take a little M from AB to make am = EC, and connect me.

∴BM=BE.∴∠BME=45。 ∴∠AME= 135。

∫CF is the bisector of the outer corner,

∴∠DCF=45。 ∴∠ECF= 135。

∴∠AME=∠ECF.

∠∠AEB+∠BAE = 90，∠AEB+CEF=90，

∴∠BAE=∠CEF.

∴△AME≌△BCF(ASA).

∴AE=EF.

(2) correct.

Proof: Take a little n on the extension line of BA, make an = ce, and then connect ne.

∴BN=BE.

∴∠N=∠FCE=45。

The quadrilateral ABCD is a square,

∴AD‖BE.

∴∠DAE=∠BEA.

∴∠NAE=∠CEF.

∴△ANE≌△ECF(ASA).

∴AE=EF.