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.

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Draw DH  perpendicular to AE

By SAS.....ΔABE  = ΔBEF

<BAE = <CBF 

<ABF  + <CBF  = 90

So <BAE + <ABF  =90

So <AGB  =90

<BAE + <DAH  = 90

<BAE + <ABF  = 90

So < DAH  = < ABF

And < AHD =< AGB  =90

So by  AAS .......ΔHAD  = ΔGBA

And by AA congruency

ΔABE ≈ ΔAGB   →  ΔBEA ≈ ΔGBA

Since AB  = the side of the square and BE  = (1/2) the side of the square

Then...... 2BE  = AB   and, by similar reasoning.....  2GB = AG

Now AG  = HD   and by the same reasoning as before 2AH  = HD  →  2AH = AG

But

AG  =AH + GH

2AH = AH + GH

So   AH  = GH

By SAS

ΔAHD = ΔGHD

And <HAD  = < HGD

But in Δ AGD, <GAD = <AGD

Therefore.......DG  = AD

But      AD  = AB

Therefore.......DG  = AB

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