Given a right triangular prism $A_1B_1C_1 - ABC$ with $\angle BAC = \frac{\pi}{2}$ and $AB = AC = AA_1$, let $G$, $E$ be the midpoints of $A_1B_1$, $CC_1$ respectively, and $D$, $F$ be variable points lying on segments $AC$, $AB$ (not including endpoints) respectively. If $GD \bot EF$, the range of the length of $DF$ is
${ \textbf{(A)}\ [\frac{1}{\sqrt{5}}, 1)\qquad\textbf{(B)}\  [\frac{1}{5}, 2)\qquad\textbf{(C)}\ [1, \sqrt{2})\qquad\textbf{(D)}} [\frac{1}{\sqrt{2}}, \sqrt{2})\qquad $
1. **Define the coordinates of the vertices:**
   Given the right triangular prism \( A_1B_1C_1 - ABC \) with \( \angle BAC = \frac{\pi}{2} \) and \( AB = AC = AA_1 \), we can place the vertices in a coordinate system as follows:
   \[
   A(0,0,0), \quad B(t,0,0), \quad C(0,t,0), \quad A_1(0,0,t), \quad B_1(t,0,t), \quad C_1(0,t,t)
   \]
   Here, \( t \) is the length of the sides \( AB \), \( AC \), and \( AA_1 \).

2. **Find the coordinates of the midpoints \( G \) and \( E \):**
   - \( G \) is the midpoint of \( A_1B_1 \):
     \[
     G = \left( \frac{0+t}{2}, \frac{0+0}{2}, \frac{t+t}{2} \right) = \left( \frac{t}{2}, 0, t \right)
     \]
   - \( E \) is the midpoint of \( CC_1 \):
     \[
     E = \left( \frac{0+0}{2}, \frac{t+t}{2}, \frac{0+t}{2} \right) = \left( 0, t, \frac{t}{2} \right)
     \]

3. **Define the coordinates of the variable points \( D \) and \( F \):**
   - Let \( D \) be a point on segment \( AC \) (not including endpoints):
     \[
     D = (0, \alpha, 0) \quad \text{where} \quad 0 < \alpha < t
     \]
   - Let \( F \) be a point on segment \( AB \) (not including endpoints):
     \[
     F = (\beta, 0, 0) \quad \text{where} \quad 0 < \beta < t
     \]

4. **Calculate the direction vectors of \( GD \) and \( EF \):**
   - The direction vector of \( GD \) is:
     \[
     \overrightarrow{GD} = D - G = (0, \alpha, 0) - \left( \frac{t}{2}, 0, t \right) = \left( -\frac{t}{2}, \alpha, -t \right)
     \]
   - The direction vector of \( EF \) is:
     \[
     \overrightarrow{EF} = F - E = (\beta, 0, 0) - \left( 0, t, \frac{t}{2} \right) = \left( \beta, -t, -\frac{t}{2} \right)
     \]

5. **Use the orthogonality condition \( GD \perp EF \):**
   - The dot product of \( \overrightarrow{GD} \) and \( \overrightarrow{EF} \) must be zero:
     \[
     \overrightarrow{GD} \cdot \overrightarrow{EF} = \left( -\frac{t}{2} \right) \beta + \alpha (-t) + (-t) \left( -\frac{t}{2} \right) = 0
     \]
     Simplifying this, we get:
     \[
     -\frac{t}{2} \beta - \alpha t + \frac{t^2}{2} = 0
     \]
     \[
     -\frac{t}{2} \beta - \alpha t + \frac{t^2}{2} = 0
     \]
     \[
     -\frac{t}{2} \beta = \alpha t - \frac{t^2}{2}
     \]
     \[
     \beta = t - 2\alpha
     \]

6. **Calculate the length of \( DF \):**
   - Using the coordinates of \( D \) and \( F \):
     \[
     DF = \sqrt{(0 - \beta)^2 + (\alpha - 0)^2 + (0 - 0)^2} = \sqrt{\beta^2 + \alpha^2}
     \]
   - Substitute \( \beta = t - 2\alpha \):
     \[
     DF = \sqrt{\alpha^2 + (t - 2\alpha)^2}
     \]
     \[
     DF = \sqrt{\alpha^2 + t^2 - 4\alpha t + 4\alpha^2}
     \]
     \[
     DF = \sqrt{5\alpha^2 - 4\alpha t + t^2}
     \]

7. **Determine the range of \( DF \):**
   - When \( \alpha = 0 \):
     \[
     DF = \sqrt{t^2} = t
     \]
   - When \( \alpha = t \):
     \[
     DF = \sqrt{5t^2 - 4t^2 + t^2} = \sqrt{2t^2} = t\sqrt{2}
     \]
   - Therefore, the range of \( DF \) is:
     \[
     t \leq DF < t\sqrt{2}
     \]
   - Normalizing by \( t \), the range is:
     \[
     1 \leq \frac{DF}{t} < \sqrt{2}
     \]

The final answer is \( \boxed{ [1, \sqrt{2}) } \)