In square $ABCD$, $E$ is the midpoint of $\overline{BC}$, and $F$ is the midpoint of $\overline{CD}$. Let $G$ be the intersection of $\overl

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

"In square ABCD, E is the midpoint of line BC, and F is the midpoint of line CD. Let G be the intersection of line AE and line BF. Prove that DG = AB."

Here's a coordinate geometry approach:

. 

Thanks, alan...I like your method better.....!!!

Thanks!