square proof using intersection of mid points

Question

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 $DG = AB$.

Answer 1

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 $$DG = AD = AB$$

Answer 2

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 $$ DG = AB.$$