Cfrgtkky

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 $DG = AB$.

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 $$ DG = AB.$$

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 $$DG = AD = AB$$