12.2.5 * In the right triangular prism $A_{1} B_{1} C_{1}-A B C$, $\angle B A C=\frac{\pi}{2}, A B=A C=A A_{1}=$ 1. Given that $G$ and $E$ are the midpoints of $A_{1} B_{1}$ and $C C_{1}$, respectively, and $D$ and $F$ are moving points on segments $A C$ and $A B$ (excluding the endpoints). If $G D \perp E F$, then the range of the length of segment $D F$ is ( ).
(A) $\left[\frac{1}{\sqrt{5}}, 1\right)$
(B) $\left[\frac{1}{5}, 2\right)$
(C) $[1, \sqrt{2})$
(D) $\left[\frac{1}{\sqrt{5}}, \sqrt{2}\right)$
Establish a Cartesian coordinate system, with $A$ as the origin, $AB$ as the $x$-axis, $AC$ as the $y$-axis, and $AA_1$ as the $z$-axis, then $E\left(0,1, \frac{1}{2}\right), G\left(\frac{1}{2}, 0,1\right)$. Let $F\left(t_{1}, 0,0\right)\left(0<t_{1}<1\right), D\left(0, t_{2}, 0\right)\left(0<t_{2}<1\right)$, then $\overrightarrow{E F}=\left(t_{1},-1,-\frac{1}{2}\right), \overrightarrow{G D}=\left(-\frac{1}{2}, t_{2},-1\right)$. Since $G D \perp E F$, we have $t_{1} + 2 t_{2} = 1$, which implies $0 < t_{2} < \frac{1}{2}$. Also, $\overrightarrow{D F}=\left(t_{1},-t_{2}, 0\right), |\overrightarrow{D F}|^{2} = \sqrt{t_{1}^{2} + t_{2}^{2}} = \sqrt{5 t_{2}^{2} - 4 t_{2} + 1} = \sqrt{5\left(t_{2} - \frac{2}{5}\right)^{2} + \frac{1}{5}}$, thus $\sqrt{\frac{1}{5}} \leqslant |\overrightarrow{D F}| < 1$. Therefore, the answer is A.