In square $ABCD$, $E$ is the midpoint of $\overline{BC}$, and $F$ is the midpoint of $\overline{CD}$. Let $G$ be the intersectio
n of $\overline{AE}$ and $\overline{BF}$. Prove that $DG = AB$.
1 answer:
Draw DH perpendicular to AE.
By the Side-Angle-Side postulate ΔABE = ΔBEF.
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