As shown in the figure, the side length of the square AOBC is 4, the inverse proportional function Y = KX passes through the center D point of the square AOBC, e is any point on the side of the AOBC,

As shown in the figure, the side length of the square AOBC is 4, the inverse proportional function Y = KX passes through the center D point of the square AOBC, e is any point on the side of the AOBC, and f is on the OB extension line. (1)∵ The center of a square is the intersection of its two diagonals.

Similarly, the side length of the square AOBC is 4,

The coordinate of the center point d of the square AOBC is (2,2).

Point ∵D is on the image of inverse proportional function y = kx,

∴k=2×2=4，

∴y=4x；

(2)CG⊥EF，CG = 12ef。 (5 points)

It is proved that connecting CE and CF makes EH∨BF cross AB at h point,

CA = CB，∠CAE=∠CBF，AE=BF，

∴△CAE≌△CBF，

∴CE=CF,∠ACE=∠BCF，

∴∠ECF=90。

AE = EH = BF，∠EGH=∠BGF，∠HEG=∠BFG，

∴△ehg≌△fbg(7 points)

∴EG=FG.

∴cg⊥ef,cg= 12ef；

(3) the intersection m is ME⊥PN in e,

∴EM∥x axis,

Let n coordinate be (a, -a+2),

∴EM=-a,NE=-a+2-2=-a，

∴ME=NE，

∴∠ PNM = 45, (9 points)

∵ Quadrilateral OPNM is an isosceles trapezoid,

∴∠ PNM =∠ NPO = 45。 (10)

∴ Set the coordinate of point P as (x, x) and substitute it into y = 4x.

∴x= 2。

∵P is a fixed point on the third quadrant of y = kx,

∴x=-2.

The coordinate of point p is (-2, -2).