So AB=AC=BC and the angle BAC=60 degrees.

In parallelogram ABCD, AB=DC, AD=BC, angle D= angle B=60 degrees.

So angle D= angle b

The other angle EAF=60 degrees

So angle CAE= angle DAF

So triangle CAE is equal to triangle DAF.

So EC=DF

So AC=CD=CF+DF=CF+CE.

2. When the angle B=45 degrees.

Easy angle D= angle ACE= angle DAC=45 degrees, angle ACD=90 degrees,

Angle EAF= angle B=45 degrees.

So simple angle DAF= angle CAE

So the triangular EAC is similar to the triangular DAF.

So there is EC/DF=AC/AD= (root number 2)/2.

So EC/(CD-CF)= (root number 2)/2.

So (AC-CF)/EC= root number 2.

3,2 shows AE/AF=AC/AD= (root number 2)/2.

So the triangle AEF is similar to the triangle ACD.

So AEF angle =90 degrees,

So the triangle AEG is similar to the triangle FCG.

So CD/AG=CF/AE

Let's ask AE

Because CF=2 and AB=4,

So DF=CD-CF=2, AC=AB=4.

Therefore, AF= root number 20 can be found in the right triangle ACF.

So in the isosceles right triangle AEF, AE= root 10 can be obtained.

So CG/AG=CF/AE

That is, CG/(AC-CG)=CF/AE.

So CG/(4-CG)=2/ root number 10.

So CG=4[ (root number 10)-2]/3.