Square and 2016

Note that $CF$ and $BE$ are perpendicular. Extend $BE$ and $CD$ to meet at $C'$ . Since $\angle CGC'=90^\circ$ , segment $CC'=4032$ would be the diameter of a circle centered at $D$ , and $G$ would be a point on the circle. Thus, $DA=DG=DC=\boxed{2016}$ .

Let D = (0,0). Then E =(0,1008), A = (0, 2016), F = (1008, 2016), B = (2016,2016), C = (2016, 0).

Slope of EB = (2016-1008)/(2016-0) = 0.5 y intercept = 1008 so equation of EB is y = 0.5x + 1008

Slope of FC = (2016 - 0)/(1008-2016) = -2. Use that fact to determine that y intercept is 4032. so equation of FC is y = -2x + 4032.

Solving equations determines intersection to be (1209.6, 1612.8).

Use distance formula to find distance DG = sqrt(1209.6 ^2 + 1612.8^2) = 2016

After Ahmad Saad's solution did not feel like posting my. However as a Varity the sketches present a solution based on similar triangles.

Consider a line going straight down from $F$ which cuts the square into 2 halves and cuts $BE$ at $I$ . By alternate angles of paraell lines, we have $\angle HFC = \angle FCB$ , $\angle FIB = \angle IBC$ and by opposite angles, $\angle FGI = \angle BGC$ . Triangles $BGC$ and $FIG$ have the same angles so they are similar.

Let a side of the square have length $x$ Since $AB = 2FB$ then by similar triangles $2FI = AE = \frac{1}{2}AD = \frac{1}{2}x$ so $FI = \frac{1}{4} x$ which implies the triangle $FIG$ is $\frac{1}{4}$ scale of triangle $FBC$ . Thereore $GC = 4FG$ and $BG = 4GI$ . Let $BG = y$ and $CG = z$ .

Consider the vertical part of $z$ . Since $\frac{1}{4} z + z = x$ , we get that $z = \frac{4}{5} x$ . Similarly consider the horizontal part of y, $EI = BI = y + \frac{1}{4}y = \frac{5}{4}y$ by similar triangles from earlier. Then $\frac{5}{4} y + \frac{1}{4}y + y = x \implies y = \frac{2}{5} x$ . That means the horizontal part of $EG = \frac{6}{4} y = \frac{3}{5} x$ .

Finally, we use Pythagoras Theorem on $DG$ . The line has a width $\frac{3}{5} x$ horizontally and a height $\frac{4}{5}x$ vertically which gives us $DG = \sqrt{\left(\frac{3}{5}x\right)^2 + \left(\frac{4}{5}x\right)^2} = x$ .

Applying this to the question, since the length of the side is 2016 then the length of $DG$ is also 2016.