If a triangle ABC, D is the mid point of side BC and `angleADB =theta " then " cot theta =`

Options

A. `(b^2+c^2)/(4Delta)`
B. `(b^2-c^2)/(4Delta)`
C. `(b^2+c^2-a^2)/(4Delta)`
D. none

Answer

To solve the problem, we need to find the value of \( \cot \theta \) in triangle \( ABC \) where \( D \) is the midpoint of side \( BC \) and \( \angle ADB = \theta \).

### Step-by-Step Solution:

1. **Draw the Triangle**:
- Start by sketching triangle \( ABC \) with points \( A \), \( B \), and \( C \). Mark point \( D \) as the midpoint of side \( BC \).

2. **Label the Sides**:
- Let \( AB = c \), \( AC = b \), and \( BC = a \). Since \( D \) is the midpoint of \( BC \), we have \( BD = DC = \frac{a}{2} \).

3. **Identify the Angles**:
- We know that \( \angle ADB = \theta \). We need to express \( \cot \theta \) in terms of the sides of the triangle.

4. **Use the MN Theorem**:
- According to the MN theorem, we can express \( \cot \theta \) as:
\[ \cot \theta = \frac{N \cot C - M \cot B}{M + N} \]
- Here, \( M = N = \frac{a}{2} \) (since \( D \) is the midpoint).

5. **Substituting Values**:
- Substitute \( M \) and \( N \) into the equation:
\[ \cot \theta = \frac{\frac{a}{2} \cot C - \frac{a}{2} \cot B}{\frac{a}{2} + \frac{a}{2}} = \frac{\frac{a}{2} (\cot C - \cot B)}{a} \]
- This simplifies to:
\[ \cot \theta = \frac{1}{2} (\cot C - \cot B) \]

6. **Expressing Cotangents**:
- Recall that:
\[ \cot C = \frac{\cos C}{\sin C}, \quad \cot B = \frac{\cos B}{\sin B} \]
- Substitute these into the equation:
\[ \cot \theta = \frac{1}{2} \left( \frac{\cos C}{\sin C} - \frac{\cos B}{\sin B} \right) \]

7. **Finding Sine and Cosine Values**:
- Use the Law of Sines:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
- This gives us:
\[ \sin B = \frac{b}{2R}, \quad \sin C = \frac{c}{2R} \]

8. **Substituting Back**:
- Substitute \( \sin B \) and \( \sin C \) back into the equation for \( \cot \theta \):
\[ \cot \theta = \frac{1}{2} \left( \frac{\cos C \cdot 2R}{c} - \frac{\cos B \cdot 2R}{b} \right) \]

9. **Final Expression**:
- After simplification, we arrive at:
\[ \cot \theta = \frac{B^2 - C^2}{4 \Delta} \]
- Where \( \Delta \) is the area of triangle \( ABC \).

### Conclusion:
Thus, the final expression for \( \cot \theta \) is:
\[ \cot \theta = \frac{B^2 - C^2}{4 \Delta} \]