square proof using intersection of mid points 
In square $ABCD$, $E$ is the midpoint of $\overline{BC}$, and $F$ is the midpoint of $\overline{CD}$. Let $G$ be the intersection of $\overline{AE}$ and $\overline{BF}$. Prove that $DG = AB$.

 A: Draw $DH$ perpendicular to $AE$.
By SAS (Side Angle Side): $$ \Delta ABE = \Delta BEF,$$ $$\angle BAE  = \angle CBF, $$ and $$\angle ABF  + \angle CBF  = 90.$$
So: $$ \angle BAE + \angle ABF  = 90.$$
Which means: $$ \angle AGB = 90, $$ $$ \angle BAE + \angle DAH  = 90, $$ and $$ \angle BAE + \angle ABF  = 90. $$
So: $$ \angle DAH  = \angle ABF, $$ and $$ \angle AHD = \angle AGB  = 90. $$
Also by AAS (Angle Angle Side): $$ \Delta HAD = \Delta GBA, $$ and by AA (Double Angle) congruency $$ \Delta ABE \approx \Delta AGB \rightarrow \Delta BEA \approx \Delta GBA. $$
Since $AB =$ the side of the square and $BE = \frac{1}{2}$ the side of the square then $2BE  = AB$ and, by similar reasoning $2GB = AG$.
Now $AG  = HD$   and by the same reasoning as before $2AH  = HD$ $\rightarrow$ $2AH = AG$.
But: $$ AG  =AH + GH, $$ and $$2AH = AH + GH.$$
So:   $$AH  = GH.$$
By SAS (Side Angle Side): $$ \Delta AHD = \Delta GHD, $$ and $$ \angle HAD  = \angle HGD. $$
But in $\Delta AGD$: $$ \angle GAD = \angle AGD. $$
Therefore: $$DG  = AD,$$ but $$AD  = AB,$$ thus $$ DG  = AB.$$
A: Note that $AGFD$ is cyclic (since $\angle ADF+\angle AGF = \pi$) and that $\angle DFA = \angle BFC=x$. Then $$\angle ADG = \angle AFB = \pi-2x$$

Also $$ \angle AGD = \angle AFD = x$$ 
So $$\angle GAD = \pi - (\pi -2x) - x =x$$ 
and thus $$DG = AD = AB$$
