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 $DG = AB$.
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
