Ok, to prove FG=GE, produce a line l || BC, such that it meets AB in T and DG in K,

Now side of the square = a

Clearly, ∆FKG ~ ∆ECG

(note : "~" is stands for similar triangles in which two triangles have all are corresponding angles equal and the very common properties of similar triangles is that the corresponding sides are all have same ratio)

now to prove the proposition given, we need to prove

∆FKG ≅ ∆ECG, So for that it is enough to show any one corresponding side is equal to that of another.

Let's try with FK and CE.

These things are easily seen :

1. AB=BC=CD=DA= a,

2. BD= BE = √2a

3. AE = √3a

4. CE = BE - BC = (√2-1)a

Now see, ∆ABE ~ ∆ GCE

so, AB/BE = GC/CE

CG = (1-1/√2)a

So DG = a - CG = a/√2 ........ (5)

Now look at ∆AFB ~ ∆GFD

So, AF/FG = AB/GD

put the value of GD from (5) in above to get

AF/FG = √2

Now, ∆ATF ~ ∆GKF

So, AF/TF = GF/KF

=> AF/FG = TF/KF

=> TF/KF = √2

So, (TF + KF)/KF = (√2 + 1)....... (6)

TF + KF = a

So (6) => FK = a/(√2+1) = (√2-1)a

which is equal to CE ... (Look at the 4th point above)

So, ∆FKG ≅ ∆ECG

=> FG = GE (proved)