To find the length of \( BD \) in the right triangle \( \triangle ABC \) with \( \angle ABC = 90^\circ \), \( AB = 6 \), and \( AC = 10 \), we start by determining \( BC \) using the Pythagorean theorem.
The Pythagorean theorem states:
\[ AB^2 + BC^2 = AC^2 \]
Substituting the known values:
\[ 6^2 + BC^2 = 10^2 \]
This simplifies to:
\[ 36 + BC^2 = 100 \]
Solving for \( BC^2 \):
\[ BC^2 = 100 - 36 = 64 \]
Taking the square root gives:
\[ BC = \sqrt{64} = 8 \]
Now we have the triangle with sides \( AB = 6 \), \( AC = 10 \), and \( BC = 8 \).
Next, we will calculate the area of triangle \( ABC \) in two ways: using the base and height, and using the sides.
- Using the sides: The area \( K \) can be calculated using the formula for the area of a triangle with the right angle:
\[ K = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 6 \times 8 = 24 \]
- Using the altitude: Let \( D \) be the foot of the altitude from \( A \) to \( BC \). The area can also be expressed as:
\[ K = \frac{1}{2} \times BC \times AD \]
Since we already computed the area \( K = 24 \):
\[ 24 = \frac{1}{2} \times 8 \times AD \]
Solving for \( AD \):
\[ 24 = 4 \times AD \quad \Rightarrow \quad AD = \frac{24}{4} = 6 \]
Now, we can apply the relationship between the segments \( BD \) and \( DC \) on hypotenuse \( BC \). Let \( BD = x \) and \( DC = y \).
With \( BD + DC = BC \):
\[ x + y = 8 \]
Using the property of similar triangles \( \triangle ABD \sim \triangle ADC \):
\[ \frac{AB}{AD} = \frac{AD}{DC} \quad \Rightarrow \quad \frac{6}{6} = \frac{6}{y} \]
This implies \( y = 6 \).
Now substituting \( y \) into the equation \( x + y = 8 \):
\[ x + 6 = 8 \quad \Rightarrow \quad x = 2 \]
Thus, the length of \( BD \) is:
\[ \boxed{2} \]