+ 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 the 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) $\left[\frac{1}{\sqrt{5}}, \sqrt{2}\right)$
Solve: Establish a rectangular 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 $F\left(t_{1}, 0,0\right)\left(0<t_{1}<1\right), E\left(0,1, \frac{1}{2}\right), G\left(\frac{1}{2}\right.$, $0,1), D\left(0, t_{2}, 0\right)\left(0<t_{2}<1\right)$. Therefore, $\overrightarrow{E F}=\left(t_{1},-1,-\frac{1}{2}\right), \overrightarrow{G D}=$ $\left(-\frac{1}{2}, t_{2},-1\right)$. Since $G D \perp E F$, we have $t_{1}+2 t_{2}=1$, which implies $0<$ $t_{2}<\frac{1}{2}$. Also, $\overrightarrow{D F}=\left(t_{1},-t_{2}, 0\right),|\overrightarrow{D F}|=\sqrt{t_{1}^{2}+t_{2}^{2}}=\sqrt{5 t_{2}^{2}-4 t_{2}+1}=$ $\sqrt{5\left(t_{2}-\frac{2}{5}\right)^{2}+\frac{1}{5}}$, thus $\sqrt{\frac{1}{5}} \leqslant|\overrightarrow{D F}|<1$. Therefore, the answer is A.