In a right triangle ABC , right angled at C, P and Q are the points of the sides CA and CB respectively , which divide these sides in the ratio `2 : 1` Prove that
`9BP^(2)=9BC^(2)+4AC^(2)`

Answer

To solve the problem, we will use the properties of right triangles and the Pythagorean theorem. Let's denote the lengths of the sides as follows:

- \( AC = b \)
- \( BC = a \)
- \( AB = c \)

Given that points \( P \) and \( Q \) divide the sides \( AC \) and \( BC \) in the ratio \( 2:1 \), we can express the segments as follows:

1. \( CP = \frac{2}{3} AC = \frac{2}{3} b \)
2. \( PA = \frac{1}{3} AC = \frac{1}{3} b \)
3. \( CQ = \frac{2}{3} BC = \frac{2}{3} a \)
4. \( QB = \frac{1}{3} BC = \frac{1}{3} a \)

We need to prove that:

\[ 9 BP^2 = 9 BC^2 + 4 AC^2 \]

### Step 1: Use the Pythagorean theorem in triangle \( CPB \)

In triangle \( CPB \), we apply the Pythagorean theorem:

\[ BP^2 = CP^2 + CB^2 \]

### Step 2: Substitute the lengths

Substituting the lengths we have:

\[ BP^2 = \left(\frac{2}{3} b\right)^2 + a^2 \]

Calculating \( CP^2 \):

\[ CP^2 = \left(\frac{2}{3} b\right)^2 = \frac{4}{9} b^2 \]

Thus, we have:

\[ BP^2 = \frac{4}{9} b^2 + a^2 \]

### Step 3: Multiply the entire equation by 9

To eliminate the fraction, we multiply the entire equation by 9:

\[ 9 BP^2 = 9 \left(\frac{4}{9} b^2 + a^2\right) \]

This simplifies to:

\[ 9 BP^2 = 4 b^2 + 9 a^2 \]

### Step 4: Rearranging the equation

We can rearrange this to match the required form:

\[ 9 BP^2 = 9 a^2 + 4 b^2 \]

### Conclusion

Thus, we have proved that:

\[ 9 BP^2 = 9 BC^2 + 4 AC^2 \]

Hence, the statement is proved.