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

Help! I need a proof!

Answer 2

Asymptote code below
[asy]
unitsize(2.5 cm);

pair A, B, C, D, E, F, G;

A = (0,0);
B = (0,1);
C = (1,1);
D = (1,0);
E = (B + C)/2;
F = (C + D)/2;
G = extension(A,E,B,F);

draw(A--B--C--D--cycle);
draw(A--E);
draw(B--F);
draw(D--G);

label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
label("$E$", E, N);
label("$F$", F, dir(0));
label("$G$", G, WSW);
[/asy]

Answer 3

Help!