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.