EG = e b-AH = 1- 1/2 = 1/2？

cos∠GED=EG/ED= 1/2？

∴∠GED=60？

Obviously, ∠DEF=∠CEF?

∴∠CEF=( 180 -∠GED)/2=60？

dg^2=ed^2-eg^2= 1- 1/4=3/4？

DG=√3/2？

DH=AB-DG=2√3-√3/2=3√3/2？

OH=OA-AH=2- 1/2=3/2？

∴D(-3/2，3√3/2)？

②∠CEF = = 60？

∴CF=ECtan60 =√3？

∴OF=OC-CF=2√3-√3=√3？

∴F(0,√3)？

E(- 1，2√3)？

Let the function expression of EF line be: y=kx b?

So: √3=b?

2√3=-k b？

∴k=-√3； b=√3？

So the function expression of ef online is: y=-√3x √3?

(3) Obviously, DF=CF=√3?

Point p is on the straight line EF. △ When △PFD is an isosceles triangle, there are three situations:?

(a)PF=DF=√3？

Let P(t,-√ 3t √ 3), then:?

PF^2=3？

(t-0)^2 (-√3t √3-√3)^2=3？

t^2 3t^2=3？

t^2=3/4？

t 1=-√3/2，t2=√3/2？

∴p 1(-√3/2,3/2√3)； P2(√3/2，-3/2 √3)？

(b) PD=DF=√3？

Let P(t, -√3t √3) and pay attention to D(-3/2, 3√3/2), then:?

PD^2=3？

(t 3/2)^2 (-√3t √3-3√3/2)^2=3？

t^2 3t 9/4 3t^2 3t 3/4=3？

4t^2 6t=0？

t 1=0，t2=-3/2？

Obviously, t 1=0 corresponds to point F. At this time, a triangle is not formed, so it is discarded.

∴P3(-3/2,5√3/2)？

(c) PD=PF？

Let P(t, -√3t √3), and pay attention to D(-3/2, 3√3/2) and F(0, √3), then:?

PD^2=PF^2？

(t 3/2)^2(-√3t √3-3√3/2)^2=(t-0)^2(-√3t √3-√3)^2？

t^2 3t 9/4 3t^2 3t 3/4=t^2 3t^2？

6t 3=0？

t=- 1/2？

∴P4(- 1/2,3√3/2)？

To sum up, there are four P's that meet the requirements, namely:?

P 1(-√3/2，3/2√3)； P2(√3/2，-3/2 √3)？

P3(-3/2，5√3/2)； P4(- 1/2，3√3/2)？