Problem 1261

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 $\boxed{25}$.