As shown in the figure (figure 1, figure 2), the quadrilateral ABCD is a square with a side length of 4, the point E is on the line BC, ∠ AEF = 90, and EF intersects the bisector CP of the square.

Solution: (1) Take a little g from AB, make AG=EC, and then connect with GE.

∴AB-AG=BC-EC，

That is BG=BE,

∴∠BGE=45，

∴∠AGE= 135。

∫CP is the bisector of the outer corner,

∴∠DCF=45，

∴∠ECF= 135，

∴∠AGE=∠ECF，

∠∠AEB+∠BAE = 90，∠AEB+∠CEF=90，

∠BAE=∠CEF，

In △AGE and △ECF, ∠ age = ∠ ecfag = EC ∠ BAE = ∠ cef,

∴△AGE≌△ECF(ASA)，

∴ae=ef；

(2)① You can prove the same reason as (1). When e is not the midpoint, AE=EF.

∴ in △ABE and △ENF, ∠ BAE = ∠ CEF ∠ B = ∠ CNF = 90 AE = EF,

∴△ABE≌△ENF(AAS)，

∴FN=BE=x，

And ∵BE=x, BC=4,

∴EC=4-x，

∴y= 12×(4-x)x，

∴y=- 12x2+2x (0