In math class, Teacher Zhang showed a problem: As shown in figure 1, the quadrilateral ABCD is a square, and the point E is the midpoint of the side BC. ∠ AEF = 90, and the parallel line CF at the outer corner of the square ∠DCG is at point F, which proves AE = EF. After thinking, Xiao Ming showed a correct method to solve the problem: take the midpoint m of AB.

(1) Xiaoying proposed: As shown in Figure 2, if "point E is the midpoint of the BC side" is changed to "point E is any point on the BC side (except B and C)", other conditions remain unchanged, then the conclusion "AE=EF" still holds. Do you think Xiaoying's point of view is correct? If it is correct, write the proof process; If not, please explain the reasons;

(2) Xiao Hua proposed that, as shown in Figure 3, point E is any point on the BC extension line (except point C), and the conclusion of "AE=EF" still holds. Do you think Xiaohua's view is correct? If it is correct, write the proof process; If not, please explain why.

Solution: (1) is correct.

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≌△ECF(ASA)，

∴AE=EF.

(2) correct.

Proof: take a little n on the extension line of BA.

Make a =CE and connect ne.

∴BN=BE，

∴∠N=∠NEC=45，

∫CF average ∠DCG,

∴∠FCE=45，

∴∠N=∠ECF，

∵ quadrilateral ABCD is a square,

∴AD∥BE，

∴∠DAE=∠BEA，

That is, ∠ DAE+90 = ∠ bea+90,

∴∠NAE=∠CEF，

∴△ANE≌△ECF(ASA)

∴AE=EF.