In a `DeltaABC`, altitude drawn from vertex A to side BC meets BC at D such that `DB=3CD`. Prove that `2AB^(2)=2AC^(2)+BC^(2)`

Answer

To prove that \( 2AB^2 = 2AC^2 + BC^2 \) in triangle \( \Delta ABC \) where the altitude from vertex \( A \) to side \( BC \) meets \( BC \) at point \( D \) such that \( DB = 3CD \), we can follow these steps:

### Step-by-Step Solution:

1. **Define the segments on \( BC \)**:
Let \( CD = x \). Then, since \( DB = 3CD \), we have:
\[ DB = 3x \]
Therefore, the total length of \( BC \) is:
\[ BC = BD + CD = 3x + x = 4x \]

2. **Identify the lengths of \( BD \) and \( CD \)**:
From the above, we can express:
\[ BD = \frac{3}{4} BC \quad \text{and} \quad CD = \frac{1}{4} BC \]

3. **Apply the Pythagorean theorem in triangles \( ABD \) and \( ACD \)**:
In triangle \( ABD \):
\[ AB^2 = AD^2 + BD^2 \]
In triangle \( ACD \):
\[ AC^2 = AD^2 + CD^2 \]

4. **Express \( AD^2 \) in terms of \( AC^2 \) and \( CD^2 \)**:
Rearranging the equation for \( AC^2 \):
\[ AD^2 = AC^2 - CD^2 \]
Substitute \( CD = \frac{1}{4} BC \):
\[ CD^2 = \left(\frac{1}{4} BC\right)^2 = \frac{1}{16} BC^2 \]
Thus,
\[ AD^2 = AC^2 - \frac{1}{16} BC^2 \]

5. **Substitute \( AD^2 \) into the equation for \( AB^2 \)**:
Now substitute \( AD^2 \) into the equation for \( AB^2 \):
\[ AB^2 = \left(AC^2 - \frac{1}{16} BC^2\right) + BD^2 \]
Substitute \( BD = \frac{3}{4} BC \):
\[ BD^2 = \left(\frac{3}{4} BC\right)^2 = \frac{9}{16} BC^2 \]
Therefore,
\[ AB^2 = AC^2 - \frac{1}{16} BC^2 + \frac{9}{16} BC^2 \]
Simplifying this gives:
\[ AB^2 = AC^2 + \frac{8}{16} BC^2 = AC^2 + \frac{1}{2} BC^2 \]

6. **Multiply the entire equation by 2**:
To prove the required statement, multiply both sides by 2:
\[ 2AB^2 = 2AC^2 + BC^2 \]

### Conclusion:
Thus, we have proven that:
\[ 2AB^2 = 2AC^2 + BC^2 \]