Given a triangle $ABC$ with integer side lengths, where $BD$ is an angle bisector of $\angle ABC$, $AD=4$, $DC=6$, and $D$ is on $AC$, compute the minimum possible perimeter of $\triangle ABC$.
1. **Apply the Angle Bisector Theorem**: The Angle Bisector Theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent sides. For triangle $ABC$ with $BD$ as the angle bisector of $\angle ABC$, we have:
   \[
   \frac{AB}{AD} = \frac{BC}{DC}
   \]
   Given $AD = 4$ and $DC = 6$, we can write:
   \[
   \frac{AB}{4} = \frac{BC}{6}
   \]
   This implies:
   \[
   6 \cdot AB = 4 \cdot BC \quad \Rightarrow \quad 3 \cdot AB = 2 \cdot BC \quad \Rightarrow \quad AB = \frac{2}{3} BC
   \]

2. **Express $AB$ in terms of $BC$**: Let $BC = x$. Then:
   \[
   AB = \frac{2}{3} x
   \]

3. **Use the Triangle Inequality**: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. For triangle $ABC$, we have:
   \[
   AB + BC > AC
   \]
   Since $AC = AD + DC = 4 + 6 = 10$, we get:
   \[
   AB + BC > 10
   \]
   Substituting $AB = \frac{2}{3} x$ and $BC = x$, we get:
   \[
   \frac{2}{3} x + x > 10 \quad \Rightarrow \quad \frac{5}{3} x > 10 \quad \Rightarrow \quad x > 6
   \]

4. **Find integer values for $BC$ and $AB$**: Since $x > 6$ and $x$ must be an integer, the smallest possible value for $x$ is 7. However, we need to check if $AB$ is also an integer:
   \[
   AB = \frac{2}{3} x
   \]
   For $AB$ to be an integer, $x$ must be a multiple of 3. The smallest multiple of 3 greater than 6 is 9. Therefore, let $x = 9$:
   \[
   BC = 9 \quad \text{and} \quad AB = \frac{2}{3} \cdot 9 = 6
   \]

5. **Calculate the perimeter**: The perimeter of triangle $ABC$ is:
   \[
   AB + BC + AC = 6 + 9 + 10 = 25
   \]

Thus, the minimum possible perimeter of $\triangle ABC$ is $\boxed{25}$.