P and Q are the mid point of the sides CA and CB respectively of a triangle ABC, right angled at C. then, find the value of 4 `(AQ^(2) + BP^(2) )`.

Answer

To solve the problem, we need to find the value of \(4 (AQ^2 + BP^2)\) where P and Q are the midpoints of sides CA and CB respectively in triangle ABC, which is right-angled at C.

### Step-by-Step Solution:

1. **Identify the Triangle and Midpoints**:
- Let triangle ABC be right-angled at C.
- Let \(A\) be at coordinates \((0, b)\), \(B\) at \((a, 0)\), and \(C\) at \((0, 0)\).
- The midpoints \(P\) and \(Q\) can be calculated as follows:
- \(P\) (midpoint of CA) = \(\left(\frac{0 + 0}{2}, \frac{b + 0}{2}\right) = \left(0, \frac{b}{2}\right)\)
- \(Q\) (midpoint of CB) = \(\left(\frac{0 + a}{2}, \frac{0 + 0}{2}\right) = \left(\frac{a}{2}, 0\right)\)

2. **Calculate the Distances \(AQ\) and \(BP\)**:
- To find \(AQ\):
\[ AQ = \sqrt{\left(0 - \frac{a}{2}\right)^2 + \left(b - 0\right)^2} = \sqrt{\left(-\frac{a}{2}\right)^2 + b^2} = \sqrt{\frac{a^2}{4} + b^2} \]
- To find \(BP\):
\[ BP = \sqrt{\left(a - 0\right)^2 + \left(0 - \frac{b}{2}\right)^2} = \sqrt{a^2 + \left(-\frac{b}{2}\right)^2} = \sqrt{a^2 + \frac{b^2}{4}} \]

3. **Square the Distances**:
- Now we need to compute \(AQ^2\) and \(BP^2\):
\[ AQ^2 = \frac{a^2}{4} + b^2 \]
\[ BP^2 = a^2 + \frac{b^2}{4} \]

4. **Combine the Squares**:
- Now, we will add \(AQ^2\) and \(BP^2\):
\[ AQ^2 + BP^2 = \left(\frac{a^2}{4} + b^2\right) + \left(a^2 + \frac{b^2}{4}\right) \]
\[ = \frac{a^2}{4} + a^2 + b^2 + \frac{b^2}{4} \]
\[ = \frac{a^2 + 4a^2 + 4b^2 + b^2}{4} = \frac{5a^2 + 5b^2}{4} = \frac{5(a^2 + b^2)}{4} \]

5. **Final Calculation**:
- Now, we need to find \(4(AQ^2 + BP^2)\):
\[ 4(AQ^2 + BP^2) = 4 \cdot \frac{5(a^2 + b^2)}{4} = 5(a^2 + b^2) \]

6. **Using Pythagorean Theorem**:
- Since triangle ABC is right-angled at C, by the Pythagorean theorem, we know:
\[ AB^2 = AC^2 + BC^2 \Rightarrow a^2 + b^2 = AB^2 \]
- Therefore:
\[ 4(AQ^2 + BP^2) = 5(AB^2) \]

### Final Answer:
Thus, the value of \(4(AQ^2 + BP^2)\) is \(5(AB^2)\).