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anonymous
 one year ago
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 $DG = AB$.
anonymous
 one year ago
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 $DG = AB$.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Help! I need a proof!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Asymptote 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(ABCDcycle); draw(AE); draw(BF); draw(DG); 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]
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