In $\vartriangle ABC, AB=AC=14 \sqrt2 , D$ is the midpoint of $CA$ and $E$ is the midpoint of $BD$. Suppose $\vartriangle CDE$ is similar to $\vartriangle ABC$. Find the length of $BD$. 
1. Given that \( \triangle ABC \) is an isosceles triangle with \( AB = AC = 14\sqrt{2} \), and \( D \) is the midpoint of \( CA \), we can determine the length of \( CD \):
   \[
   CD = \frac{1}{2} AC = \frac{1}{2} \times 14\sqrt{2} = 7\sqrt{2}
   \]

2. Let \( BD = x \). Since \( E \) is the midpoint of \( BD \), the length of \( DE \) is:
   \[
   DE = \frac{1}{2} BD = \frac{x}{2}
   \]

3. Given that \( \triangle CDE \) is similar to \( \triangle ABC \), the ratio of corresponding sides must be equal. Therefore:
   \[
   \frac{CD}{DE} = \frac{AB}{BC}
   \]

4. Substituting the known lengths:
   \[
   \frac{7\sqrt{2}}{\frac{x}{2}} = \frac{14\sqrt{2}}{BC}
   \]

5. Simplifying the left-hand side:
   \[
   \frac{7\sqrt{2}}{\frac{x}{2}} = \frac{7\sqrt{2} \times 2}{x} = \frac{14\sqrt{2}}{x}
   \]

6. Equating the ratios:
   \[
   \frac{14\sqrt{2}}{x} = \frac{14\sqrt{2}}{BC}
   \]

7. From the equation above, it follows that:
   \[
   x = BC
   \]

8. Since \( BD = BC \), \( \triangle BDC \) is an isosceles triangle. Given that \( \triangle ABC \sim \triangle BDC \), we use the similarity ratio:
   \[
   \frac{BD}{DC} = \frac{AB}{BC}
   \]

9. Substituting the known lengths:
   \[
   \frac{x}{7\sqrt{2}} = \frac{14\sqrt{2}}{x}
   \]

10. Cross-multiplying to solve for \( x \):
    \[
    x^2 = 14\sqrt{2} \times 7\sqrt{2}
    \]

11. Simplifying the right-hand side:
    \[
    x^2 = 14 \times 7 \times 2 = 196 \times 2 = 392
    \]

12. Taking the square root of both sides:
    \[
    x = \sqrt{392} = 14
    \]

The final answer is \( \boxed{14} \).