$\vartriangle ABC$ has side lengths $AB = 4$ and $AC = 9$. Angle bisector $AD$ bisects angle $A$ and intersects $BC$ at $D$. Let $k$ be the ratio $\frac{BD}{AB}$ . Given that the length $AD$ is an integer, find the sum of all possible $k^2$
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1. Given the triangle \( \triangle ABC \) with side lengths \( AB = 4 \) and \( AC = 9 \), and the angle bisector \( AD \) intersecting \( BC \) at \( D \), we need to find the sum of all possible values of \( k^2 \) where \( k = \frac{BD}{AB} \).

2. By the Angle Bisector Theorem, we have:
   \[
   \frac{BD}{DC} = \frac{AB}{AC} = \frac{4}{9}
   \]
   Let \( BD = 4k \) and \( DC = 9k \). Therefore, the length of \( BC \) is:
   \[
   BC = BD + DC = 4k + 9k = 13k
   \]

3. Applying the triangle inequality:
   \[
   AB + BC > AC \implies 4 + 13k > 9 \implies 13k > 5 \implies k > \frac{5}{13}
   \]
   \[
   AB + AC > BC \implies 4 + 9 > 13k \implies 13 > 13k \implies k < 1
   \]
   Thus, we have:
   \[
   \frac{5}{13} < k < 1
   \]

4. Using the extended angle bisector theorem, which states:
   \[
   AD^2 + BD \cdot DC = AB \cdot AC
   \]
   Substituting the known values:
   \[
   AD^2 + (4k)(9k) = 4 \cdot 9
   \]
   \[
   AD^2 + 36k^2 = 36
   \]
   \[
   AD^2 = 36 - 36k^2
   \]
   \[
   AD^2 = 36(1 - k^2)
   \]
   \[
   AD = 6\sqrt{1 - k^2}
   \]

5. Since \( AD \) is an integer, let \( AD = 6m \) where \( m \) is a rational number. Then:
   \[
   6m = 6\sqrt{1 - k^2}
   \]
   \[
   m = \sqrt{1 - k^2}
   \]
   Given \( \frac{5}{13} < k < 1 \), we have \( 1 > k^2 > \frac{25}{169} \). Therefore:
   \[
   1 > 1 - k^2 > \frac{144}{169}
   \]
   \[
   1 > \sqrt{1 - k^2} > \frac{12}{13}
   \]

6. The possible values for \( m \) in the range \( \frac{12}{13} > m > 0 \) that are rational and make \( AD \) an integer are:
   \[
   m = \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \frac{1}{3}, \frac{1}{6}
   \]

7. For each \( m \):
   \[
   \sqrt{1 - k^2} = \frac{5}{6}, \frac{2}{3}, \frac{1}{2}, \frac{1}{3}, \frac{1}{6}
   \]
   \[
   1 - k^2 = \left(\frac{5}{6}\right)^2, \left(\frac{2}{3}\right)^2, \left(\frac{1}{2}\right)^2, \left(\frac{1}{3}\right)^2, \left(\frac{1}{6}\right)^2
   \]
   \[
   1 - k^2 = \frac{25}{36}, \frac{4}{9}, \frac{1}{4}, \frac{1}{9}, \frac{1}{36}
   \]
   \[
   k^2 = 1 - \frac{25}{36}, 1 - \frac{4}{9}, 1 - \frac{1}{4}, 1 - \frac{1}{9}, 1 - \frac{1}{36}
   \]
   \[
   k^2 = \frac{11}{36}, \frac{5}{9}, \frac{3}{4}, \frac{8}{9}, \frac{35}{36}
   \]

8. Summing all possible values of \( k^2 \):
   \[
   \frac{11}{36} + \frac{5}{9} + \frac{3}{4} + \frac{8}{9} + \frac{35}{36}
   \]
   Converting to a common denominator:
   \[
   \frac{11}{36} + \frac{20}{36} + \frac{27}{36} + \frac{32}{36} + \frac{35}{36} = \frac{125}{36}
   \]

The final answer is \( \boxed{\frac{125}{36}} \).