P and Q are the mid-points of the sides CA and CB respectively of a `DeltaABC`, right angled at C, then find :
(i) `4AC^(2)+BC^(2)`
(ii) `4BC^(2)+AC^(2)`
(iii) `4(AQ^(2)+BP^(2))`

Options

A. `{:("(i)","(ii)","(iii)"),(4AQ^(2),4BP^(2),5AB^(2)):}`
B. `{:("(i)","(ii)","(iii)"),(5AQ^(2),5BP^(2),4AB^(2)):}`
C. `{:("(i)","(ii)","(iii)"),(4AQ^(2),5BP^(2),5AB^(2)):}`
D. `{:("(i)","(ii)","(iii)"),(5AQ^(2),4BP^(2),4AB^(2)):}`

Answer

To solve the problem step by step, we will analyze each part of the question separately.

### Given:
- Triangle \( ABC \) is right-angled at \( C \).
- \( P \) and \( Q \) are the midpoints of sides \( CA \) and \( CB \) respectively.

### (i) Find \( 4AC^2 + BC^2 \)

1. **Identify the right triangle**: In triangle \( ACQ \), since \( C \) is the right angle, we can apply the Pythagorean theorem.

\[ AQ^2 = AC^2 + CQ^2 \]

2. **Express \( AC^2 \)**: Rearranging the equation gives us:

\[ AC^2 = AQ^2 - CQ^2 \]

3. **Express \( BC \)**: Since \( Q \) is the midpoint of \( CB \), we can express \( BC \) in terms of \( CQ \):

\[ BC = 2CQ \]

4. **Substitute \( BC \)**: Now substituting \( BC \) into the equation:

\[ 4AC^2 + BC^2 = 4(AQ^2 - CQ^2) + (2CQ)^2 \]

5. **Simplify the equation**:

\[ = 4AQ^2 - 4CQ^2 + 4CQ^2 \]

The \( -4CQ^2 \) and \( +4CQ^2 \) cancel out:

\[ = 4AQ^2 \]

### **Final Result for (i)**:
\[ 4AC^2 + BC^2 = 4AQ^2 \]