ABC is an equilateral triangle, P is a point in BC such that BP : PC =2 : 1.
Prove that : `9 AP^(2) =7 AB^(2)`

Answer

To prove that \( 9 AP^2 = 7 AB^2 \) in the given equilateral triangle \( ABC \) with point \( P \) on side \( BC \) such that \( BP : PC = 2 : 1 \), we can follow these steps:

### Step 1: Define the lengths
Let \( BP = 2x \) and \( PC = x \). Therefore, the total length of \( BC \) is:
\[ BC = BP + PC = 2x + x = 3x \]

### Step 2: Identify the lengths of the sides of the triangle
Since \( ABC \) is an equilateral triangle, all sides are equal. Thus:
\[ AB = AC = BC = 3x \]

### Step 3: Draw the altitude from point \( A \) to side \( BC \)
Let \( M \) be the midpoint of \( BC \). Since \( M \) is the midpoint, we have:
\[ BM = MC = \frac{BC}{2} = \frac{3x}{2} = 1.5x \]

### Step 4: Determine the lengths of segments
Since \( P \) divides \( BC \) in the ratio \( 2:1 \), we can find the lengths:
\[ MP = BM - BP = 1.5x - 2x = -0.5x \quad \text{(not applicable, we need to find the correct segment)} \]
Instead, we can find \( PM \):
\[ PM = PC - MC = x - 1.5x = -0.5x \quad \text{(this indicates we need to adjust our segments)} \]
We can also find \( PM \) as:
\[ PM = 1.5x - x = 0.5x \]

### Step 5: Apply the Pythagorean theorem in triangles \( ABM \) and \( AMP \)
For triangle \( ABM \):
\[ AB^2 = AM^2 + BM^2 \]
Substituting the known values:
\[ (3x)^2 = AM^2 + (1.5x)^2 \]
\[ 9x^2 = AM^2 + 2.25x^2 \]
\[ AM^2 = 9x^2 - 2.25x^2 = 6.75x^2 \]

For triangle \( AMP \):
\[ AP^2 = AM^2 + MP^2 \]
Substituting the known values:
\[ AP^2 = AM^2 + (0.5x)^2 \]
\[ AP^2 = 6.75x^2 + 0.25x^2 = 7x^2 \]

### Step 6: Relate \( AP^2 \) to \( AB^2 \)
Now, substituting back into the equation:
\[ 9AP^2 = 9(7x^2) = 63x^2 \]
And for \( AB^2 \):
\[ AB^2 = 9x^2 \]
Thus:
\[ 9AP^2 = 7AB^2 \]
\[ 63x^2 = 7(9x^2) = 63x^2 \]

### Conclusion
Therefore, we have proved that:
\[ 9AP^2 = 7AB^2 \]