China Mathematical Competition

We establish a coordinate system with point $A$ as the origin, line $AB$ as the $x$-axis, $AC$ the $y$-axis and $A{A}_1$ the $z$-axis. Then we have $F(t_1,0,0)$ ($0<t_1<1$), $E(0,1,\frac{1}{2})$, $G(\frac{1}{2},0,1)$, $D(0,t_2,0)$ ($0<t_2<1$). Therefore $\overrightarrow{EF}=(t_1,-1,-\frac{1}{2})$, $\overrightarrow{GD}=(-\frac{1}{2},t_2,-1)$. Since $GD\perp EF$, we get $t_1+2t_2=1$. Then $0<t_2<\frac{1}{2}$. Furthermore, $\overrightarrow{DF}=(t_1,-t_2,0)$, $$|\overrightarrow{DF}|=\sqrt{t_1^2+t_2^2}=\sqrt{5t_2^2-4t_2+1}=\sqrt{5{\left(t_2-\frac{2}{5}\right)}^2+\frac{1}{5}}$$ We obtain $\sqrt{\frac{1}{5}}\le |\overrightarrow{DF}|<1$. Answer: A.