If ABC is a right angled triangle at B and M, N are the mid-points of AB and BC, then `4(AN^2+CM^2)` is equal to :

Options

A. `4AC^2`
B. `6AC^2`
C. `5AC^2`
D. `5/4AC^2`

Answer

To solve the problem, we need to find the value of \( 4(AN^2 + CM^2) \) in terms of \( AC^2 \) for the right-angled triangle \( ABC \) with the right angle at \( B \).

### Step-by-Step Solution:

1. **Identify the Triangle and Midpoints**:
- Let \( A \) be at \( (0, 0) \), \( B \) at \( (a, 0) \), and \( C \) at \( (a, b) \).
- The midpoints \( M \) and \( N \) are:
- \( M \) (midpoint of \( AB \)): \( M = \left( \frac{0 + a}{2}, \frac{0 + 0}{2} \right) = \left( \frac{a}{2}, 0 \right) \)
- \( N \) (midpoint of \( BC \)): \( N = \left( \frac{a + a}{2}, \frac{0 + b}{2} \right) = \left( a, \frac{b}{2} \right) \)

2. **Calculate \( AN^2 \)**:
- The distance \( AN \) is given by:
\[ AN = \sqrt{(a - 0)^2 + \left(\frac{b}{2} - 0\right)^2} = \sqrt{a^2 + \left(\frac{b}{2}\right)^2} = \sqrt{a^2 + \frac{b^2}{4}} \]
- Therefore, \( AN^2 = a^2 + \frac{b^2}{4} \).

3. **Calculate \( CM^2 \)**:
- The distance \( CM \) is given by:
\[ CM = \sqrt{\left(\frac{a}{2} - a\right)^2 + (0 - b)^2} = \sqrt{\left(-\frac{a}{2}\right)^2 + (-b)^2} = \sqrt{\frac{a^2}{4} + b^2} \]
- Therefore, \( CM^2 = \frac{a^2}{4} + b^2 \).

4. **Combine \( AN^2 \) and \( CM^2 \)**:
- Now, we can add \( AN^2 \) and \( CM^2 \):
\[ AN^2 + CM^2 = \left(a^2 + \frac{b^2}{4}\right) + \left(\frac{a^2}{4} + b^2\right) \]
- Simplifying this gives:
\[ AN^2 + CM^2 = a^2 + \frac{a^2}{4} + \frac{b^2}{4} + b^2 = a^2 + \frac{a^2}{4} + \frac{5b^2}{4} \]
- Combine like terms:
\[ AN^2 + CM^2 = \frac{5a^2}{4} + \frac{5b^2}{4} = \frac{5(a^2 + b^2)}{4} \]

5. **Relate to \( AC^2 \)**:
- Since \( AC^2 = a^2 + b^2 \), we can substitute:
\[ AN^2 + CM^2 = \frac{5}{4} AC^2 \]

6. **Final Calculation**:
- Now, we multiply by 4 to find \( 4(AN^2 + CM^2) \):
\[ 4(AN^2 + CM^2) = 4 \cdot \frac{5}{4} AC^2 = 5 AC^2 \]

### Conclusion:
Thus, the value of \( 4(AN^2 + CM^2) \) is \( 5 AC^2 \).