In right-angled `DeltaABC, D and E ` trisect `BC`. Find the value of 8 `((AE)/(AC))^2 - 5 ((AD)/(AC))^2`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( 8 \left( \frac{AE}{AC} \right)^2 - 5 \left( \frac{AD}{AC} \right)^2 \) in the right-angled triangle \( \Delta ABC \) where points \( D \) and \( E \) trisect \( BC \).

### Step-by-Step Solution:

1. **Identify the Triangle and Trisection Points**:
- Let \( BC \) be trisected by points \( D \) and \( E \). This means \( BD = DE = EC \).
- Let \( BD = DE = EC = x \). Therefore, \( BC = 3x \).

2. **Apply the Pythagorean Theorem**:
- In triangle \( ABD \):
\[ AD^2 = AB^2 + BD^2 = AB^2 + x^2 \quad \text{(Equation 1)} \]
- In triangle \( ABE \):
\[ AE^2 = AB^2 + BE^2 = AB^2 + (BD + DE)^2 = AB^2 + (2x)^2 = AB^2 + 4x^2 \quad \text{(Equation 2)} \]
- In triangle \( ABC \):
\[ AC^2 = AB^2 + BC^2 = AB^2 + (3x)^2 = AB^2 + 9x^2 \quad \text{(Equation 3)} \]

3. **Express \( \frac{AE}{AC} \) and \( \frac{AD}{AC} \)**:
- From Equations 2 and 3, we have:
\[ \frac{AE^2}{AC^2} = \frac{AB^2 + 4x^2}{AB^2 + 9x^2} \]
- From Equations 1 and 3, we have:
\[ \frac{AD^2}{AC^2} = \frac{AB^2 + x^2}{AB^2 + 9x^2} \]

4. **Substitute into the Expression**:
- Substitute the expressions for \( \frac{AE^2}{AC^2} \) and \( \frac{AD^2}{AC^2} \) into the original expression:
\[ 8 \left( \frac{AE^2}{AC^2} \right) - 5 \left( \frac{AD^2}{AC^2} \right) = 8 \cdot \frac{AB^2 + 4x^2}{AB^2 + 9x^2} - 5 \cdot \frac{AB^2 + x^2}{AB^2 + 9x^2} \]

5. **Combine the Terms**:
- Combine the fractions:
\[ = \frac{8(AB^2 + 4x^2) - 5(AB^2 + x^2)}{AB^2 + 9x^2} \]
- Simplifying the numerator:
\[ = \frac{8AB^2 + 32x^2 - 5AB^2 - 5x^2}{AB^2 + 9x^2} = \frac{(8 - 5)AB^2 + (32 - 5)x^2}{AB^2 + 9x^2} \]
\[ = \frac{3AB^2 + 27x^2}{AB^2 + 9x^2} \]

6. **Factor Out Common Terms**:
- Factor out \( 3 \) from the numerator:
\[ = \frac{3(AB^2 + 9x^2)}{AB^2 + 9x^2} = 3 \]

### Final Answer:
Thus, the value of \( 8 \left( \frac{AE}{AC} \right)^2 - 5 \left( \frac{AD}{AC} \right)^2 \) is \( \boxed{3} \).