Geometry Proof Concerning Lengths Within a Square 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: Let us assume that $AB=2$. Then we have $BE=1$, and since $AE\perp BF$ we also have $AE=\sqrt{5}$ by the Pythagorean theorem. Since $AG\cdot AE = AB^2$ ($AGB$ and $ABE$ are similar triangles) we have $AG=\frac{4}{\sqrt{5}}$. Since $\cos\widehat{GAD}=\frac{1}{\sqrt{5}}$ we may easily compute $DG^2$ through the cosine theorem
$$DG^2 = AD^2+AG^2-2\cdot AD\cdot AG\cdot\cos\widehat{GAD} = 4+\frac{16}{5}-\frac{16}{\sqrt{5}}\cdot\frac{1}{\sqrt{5}}$$
and deduce that $DG^2=AD^2=AB^2$ as wanted.

With a simpler approach, it is enough to notice that $D$, the midpoint of $AG$ and the midpoint of $AB$ are collinear: they all lie on the parallel to $BF$ through the midpoint of $AB$. In particular the median through $D$ in $AGD$ is also a perpendicular bisector and $DG=DA=AB$ follows.

Let us state it clearly: there is a single parallel to $BF$ through the midpoint of $AB$. It goes through the midpoint of $AG$ by Thales' theorem and through $D$ by symmetry with respect to the center of the square (or by translation). Once we get that the red line is a median, from $BF\perp AE$ we also get that it is a height for $ADG$, hence $AD=DG$ by SAS.
A: 
$AE$ and $BF$ are perpendicular. You can prove it by performing a $90^{\circ}$ counter-clockwise rotation around the center of the square, which sends triangle $BCF$ to triangle $ABE$, which means that $BF$ is mapped to $AE$ so $BF=AE$ and $BF \perp AE$. Alternatively, directly show that triangles $ABE$ and  $BCF$ are congruent, i.e. $AB=BC, \, BE=CF$ and $\angle \, B = \angle \, C = 90^{\circ}$. Then by a short angle chase you get  $AE \perp BF$
Extend $BF$ until it intersects line $AD$ at point $H$. Since $F$ is the midpoint of $CD$, one concludes that $HD = BC = AD$ so $D$ is the midpoint of $AH$. Triangle $AGH$ is right-angled, where $\angle \, AGH = 90^{\circ}$, because $AE \perp BF$, so the median $DG$ is one half of the hypotenuse $AH$. Hence $DG=AD=AB$.
