Teacher Zhang asked a question in the math class.

In math class, teacher Zhang asks questions; As shown in figure 1, the quadrilateral ABCD is a square, the point e is the midpoint of the side BC, ∠ AEF = 90, and the bisector CF of the outer corner of the square ∠DCG is f, which proves AE = EF. After thinking, Xiao Ming showed the correct way to solve the problem; If the midpoint m of AB is connected to ME, then AM=EC, which is easy to prove △ AME △ ECF, so AE=EF. On this basis, the students did further research;

(1) Xiaoying proposed: As shown in Figure 2, if "point E is the midpoint of BC" is changed to "point E is any point on BC (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) Xiaohua proposed that, as shown in Figure 3, point E is a point on the BC extension line (except point C), and the conclusion of "AE=EF" still holds, with other conditions unchanged. 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≌△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.