Probably a way easier way to do this with Geometry...but....I didn't see it...so.....

By SAS, triangles BCF and ABE are congruent

Therefore angle EAB = angle FBC

And angle DAE = angle BEA

But angle DAE + angle EAB = 90

But angle FBC + angle FBA = 90

So angle BEA = angle FBA

So...by AA congruence....triangles BEG and ABG are similar

But angle BGA = angle EGB...so....each must = 90

So triangles BEG an ABG are right triangles

And AB = 2BE

So AG = 2 BG

So AB^2 = AG^2 + BG^2

AB^2 = (2BG)^2 + BG^2

AB^2 = 5BG^2

So AB = sqrt(5)BG = BC

And (1/2)AB = [sqrt(5)/2] BG

cos angle ABG = BG/ AB = BG / sqrt(5)BG = 1/ sqrt(5)

sin angle ABG =2BG/sqrt (5)BG = 2/sqrt(5)

And angle ABG = angle CFB

So sin angle ABG = sin angle CFB

But angle CFB and angle GFD are supplementary so their sines are equal

So....sin ABG = sin GFD

And GFD is obtuse...so

So... cos GFD = -sqrt [ 1 - sin^2(ABG) ] = -sqrt [ 1 - sin^2(GFD) ] =

- sqrt [ 1 - (2/sqrt(5))^2 ] = -sqrt [ 1 - 4/5] = -sqrt (1 /5) = -1/sqrt (5)

And

Triangles ABE and BCF are congruent by SAS

EA = FB BC = AB and BE = CF = (1/2)AB = FD

EA^2 = BC^2 + BE^2 .... so.....

FB^2 = AB^2 + (AB/2)^2

FB^2 = 5BG^2 + AB^2/4

FB^2 = 5BG^2 + (5/4)BG^2

FB = BGsqrt (5 + 5/4)

FB = BGsqrt [ 25/4] = (5/2)BG

And

FG = FB - BG = (5/2)BG - BG = (3/2)BG

So using the Law of Cosines

GD^2 = FG^2 + FD^2 - 2(FG)(FD)cosGFD

GD^2 = FG^2 + (AB/2)^2 - 2(FG)(AB/2)cosGFD

GD^2 = (9/4)BG^2 + (5/4)BG^2 - 2 (3/2)BG * ( sqrt(5)BG/2) * [ - (1/sqrt(5) ) ]

GD^2 = (14/4)BG^2 + (3/2)BG^2

GD^2 = (7/2)BG^2 + (3/2) BG^2

GD^2 = (10/2)BG^2

GD^2 = 5BG^2

GD = sqrt(5)BG

DG = sqrt(5)BG = AB