Question

In a right triangle △ABC, ∠ABC=90∘, AB=6, and AC=10. Point D lies on side BC,

and AD is the altitude from vertex A to the hypotenuse BC. Find the length of BD.

Answer 1

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.

1. **Using the sides**:
The area \( K \) can be calculated using the formula for the area of a triangle with the right angle:

\[
K = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 6 \times 8 = 24
\]

2. **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:

\[
\boxed{2}
\]