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

Since ABCD is a square
BE = FD
BC = AB
Angle ABE = angle BCF
With these, triangle ABE = triangle BCF
And
angle BEA = angle CFB by CPCTC
And with this
DG = AB