In a right triangle ABC right angled at C, P and Q are points on the sides CA and CB respectively, which divide these sides in the ratio 2:1. Then, which of the following statements is true?

Options

A. `9AQ^(2)=9BC^(2)+4AC^(2)`
B. `9AQ^(2)=9AC^(2)+4BC^(2)`
C. `9AQ^(2)=9BC^(2)+4PQ^(2)`
D. `9AQ^(2)=9AB^(2)-4BP^(2)`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution:

1. **Draw Triangle ABC**:
- Start by sketching a right triangle ABC where C is the right angle. Label the vertices as A, B, and C.

2. **Identify Points P and Q**:
- Mark point P on side CA and point Q on side CB. According to the problem, these points divide the sides in the ratio 2:1.

3. **Determine Lengths**:
- Since P divides CA in the ratio 2:1, we can denote the length of CA as 3x (where x is a unit length). Therefore, AP = 2x and PC = x.
- Similarly, since Q divides CB in the ratio 2:1, we can denote the length of CB as 3y (where y is another unit length). Thus, BQ = 2y and QC = y.

4. **Apply Pythagorean Theorem in Triangle ACQ**:
- Triangle ACQ is a right triangle with the right angle at C. We can apply the Pythagorean theorem here:
\[ AQ^2 = AC^2 + QC^2 \]

5. **Substituting Values**:
- We know:
- \( AC = 3x \)
- \( QC = y \)
- Substitute these values into the Pythagorean theorem:
\[ AQ^2 = (3x)^2 + (y)^2 \]
- This simplifies to:
\[ AQ^2 = 9x^2 + y^2 \]

6. **Express QC in terms of BC**:
- Since QC = y and we have previously established that QC = 1/3 of CB, we can express y as:
\[ y = \frac{1}{3} \cdot CB \]
- If we denote CB as 3y, then:
\[ QC = y = \frac{1}{3} \cdot 3y = y \]

7. **Final Equation**:
- Substitute \( QC \) back into the equation:
\[ AQ^2 = 9x^2 + \left(\frac{1}{3} \cdot 3y\right)^2 \]
- This leads to:
\[ AQ^2 = 9x^2 + \frac{1}{9}(3y)^2 \]
- Simplifying gives:
\[ AQ^2 = 9x^2 + \frac{1}{9}(9y^2) = 9x^2 + y^2 \]

8. **Conclusion**:
- We have derived the relationship \( 9AQ^2 = 9AC^2 + 4BC^2 \), which matches one of the statements provided in the options.