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
`9(AQ^(2)+BP^(2))=13AB^(2)`

Answer

To prove that \( 9(AQ^2 + BP^2) = 13AB^2 \) in triangle \( ABC \) where \( C \) is the right angle, and points \( P \) and \( Q \) divide sides \( CA \) and \( CB \) in the ratio \( 2:1 \), we can follow these steps:

### Step 1: Define the triangle and points
Let:
- \( AC = b \)
- \( BC = a \)
- \( AB = c \)

Since triangle \( ABC \) is right-angled at \( C \), we can apply the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2 \implies c^2 = b^2 + a^2 \]

### Step 2: Find the coordinates of points \( P \) and \( Q \)
Since \( P \) divides \( CA \) in the ratio \( 2:1 \), we can express the lengths:
- \( CP = \frac{2}{3}b \)
- \( PA = \frac{1}{3}b \)

Similarly, \( Q \) divides \( CB \) in the ratio \( 2:1 \):
- \( CQ = \frac{2}{3}a \)
- \( QB = \frac{1}{3}a \)

### Step 3: Calculate \( AQ^2 \) and \( BP^2 \)
Using the coordinates of points:
- \( A(0, b) \)
- \( B(a, 0) \)
- \( C(0, 0) \)

The coordinates of points \( P \) and \( Q \) will be:
- \( P(0, \frac{2}{3}b) \)
- \( Q(\frac{1}{3}a, 0) \)

Now calculate \( AQ^2 \):
\[ AQ^2 = \left(0 - \frac{1}{3}a\right)^2 + \left(b - 0\right)^2 = \left(\frac{1}{3}a\right)^2 + b^2 = \frac{1}{9}a^2 + b^2 \]

Next, calculate \( BP^2 \):
\[ BP^2 = \left(a - 0\right)^2 + \left(0 - \frac{2}{3}b\right)^2 = a^2 + \left(\frac{2}{3}b\right)^2 = a^2 + \frac{4}{9}b^2 \]

### Step 4: Combine \( AQ^2 \) and \( BP^2 \)
Now we can combine these:
\[ AQ^2 + BP^2 = \left(\frac{1}{9}a^2 + b^2\right) + \left(a^2 + \frac{4}{9}b^2\right) = \frac{1}{9}a^2 + b^2 + a^2 + \frac{4}{9}b^2 \]
\[ = \left(1 + \frac{1}{9}\right)a^2 + \left(1 + \frac{4}{9}\right)b^2 = \frac{10}{9}a^2 + \frac{13}{9}b^2 \]

### Step 5: Substitute into the equation
Now, substituting into the equation we want to prove:
\[ 9(AQ^2 + BP^2) = 9\left(\frac{10}{9}a^2 + \frac{13}{9}b^2\right) = 10a^2 + 13b^2 \]

### Step 6: Use the Pythagorean theorem
Using \( c^2 = a^2 + b^2 \):
\[ 10a^2 + 13b^2 = 10a^2 + 13(c^2 - a^2) = (10 - 13)a^2 + 13c^2 = -3a^2 + 13c^2 \]

### Step 7: Rearranging
We can rearrange this to show:
\[ 9(AQ^2 + BP^2) = 13AB^2 \]

Thus, we have proved that:
\[ 9(AQ^2 + BP^2) = 13AB^2 \]