D1f/sin (90-θ) = BF/sin30 = BD1/sin (60+θ), namely:

D 1F=BCcosθ/sin(60+θ)，BF=BC/sin(60+θ)

e 1F = e 1d 1-d 1F = BC[2/√3

-cosθ/sin(60+θ)]

△E 1EF is similar to △GAF, and FG/E 1F=AF/BF is obtained.

= & gt

E 1G/E 1F=BC/BF

Therefore: e1g = BC * e1f/BF = BC [4sin (60+θ)/√ 3.

-2cosθ]

The sum angle formula can be simplified as: E 1G=BC*2sinθ/√3.

When θ = 45, E 1G/BC=√2/√3.

(2) Yi Zheng: n is the midpoint of E 1D 1. If E 1D 1= 1, then nd1= bn = e1d/kloc-.

△D2MN is similar to △D 1BN, and it is: d2m/d1b = d2n/d1n = (bd2-bn)/(1/2) = (√ 3/2).

- 1/2)/( 1/2)=√3

- 1

D2M=√3/2*(√3)

- 1)=3/2-√3/2

E2M = e 1d 1-D2M = 1-(3/2-√3/2)=√3/2

- 1/2

So: E2M/BN=√3

- 1

It is also proved here that ME2=MN.