If in a `triangleABC, A=pi/3 and AD` is the median, then

Options

A. `2AD^(2)=b^(2)+c^(2)+bc`
B. `4AD^(2)=b^(2)+c^(2)+bc`
C. `6AD^(2)=b^(2)+c^(2)+bc`
D. none of these

Answer

To solve the problem step by step, we will use the properties of triangles and the cosine rule.

### Step-by-Step Solution:

1. **Given Information**:
- In triangle \( ABC \), \( A = \frac{\pi}{3} \) (which is 60 degrees).
- \( AD \) is the median from vertex \( A \) to side \( BC \).

2. **Setting Up the Triangle**:
- Let \( AB = c \), \( AC = b \), and \( BC = a \).
- Since \( AD \) is the median, it divides \( BC \) into two equal parts. Thus, \( BD = DC = \frac{a}{2} \).

3. **Applying the Cosine Rule**:
- According to the cosine rule, we have:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
- Substituting \( A = \frac{\pi}{3} \) into the equation:
\[ \cos \frac{\pi}{3} = \frac{b^2 + c^2 - a^2}{2bc} \]
- Since \( \cos \frac{\pi}{3} = \frac{1}{2} \), we can set up the equation:
\[ \frac{1}{2} = \frac{b^2 + c^2 - a^2}{2bc} \]

4. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 1 \cdot 2bc = b^2 + c^2 - a^2 \]
- Rearranging this, we get:
\[ b^2 + c^2 - a^2 = bc \quad \text{(Equation 1)} \]

5. **Using the Cosine Rule in Triangle ABD**:
- In triangle \( ABD \), we apply the cosine rule again:
\[ \cos B = \frac{c^2 + \left(\frac{a}{2}\right)^2 - AD^2}{b \cdot \frac{a}{2}} \]
- Rearranging gives:
\[ AD^2 = c^2 + \left(\frac{a}{2}\right)^2 - b \cdot \frac{a}{2} \cdot \cos B \]

6. **Substituting and Simplifying**:
- We know \( \cos B = \frac{c^2 + b^2 - a^2}{2bc} \).
- Substituting this into our equation for \( AD^2 \) and simplifying will lead us to:
\[ 4AD^2 = 2c^2 + 2b^2 - a^2 \]

7. **Final Substitution**:
- Using Equation 1, substitute \( a^2 \) from \( b^2 + c^2 - bc \) into the equation:
\[ 4AD^2 = 2c^2 + 2b^2 - (b^2 + c^2 - bc) \]
- This simplifies to:
\[ 4AD^2 = b^2 + c^2 + bc \]

8. **Conclusion**:
- Therefore, the final result is:
\[ 4AD^2 = b^2 + c^2 + bc \]