In `Delta ABC, (a + b+ c) (b + c -a) = kbc` if

Options

A. `k lt 0`
B. `k gt 0`
C. `0 lt k lt 4`
D. `k lt 4`

Explanation

To solve the problem step by step, we will analyze the equation given in the triangle and derive the conditions for \( k \).

### Step 1: Start with the given equation
We have the equation:
\[ (a + b + c)(b + c - a) = kbc \]

### Step 2: Expand the left-hand side
Expanding the left-hand side:
\[ a(b + c) + b(b + c) + c(b + c) - a^2 - ab - ac = kbc \]
This simplifies to:
\[ b^2 + c^2 + 2bc - a^2 = kbc \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ b^2 + c^2 - a^2 + 2bc - kbc = 0 \]

### Step 4: Use the cosine rule
Using the cosine rule, we know:
\[ b^2 + c^2 - a^2 = 2bc \cos A \]
Substituting this into our equation:
\[ 2bc \cos A + 2bc - kbc = 0 \]

### Step 5: Factor out \( bc \)
Factoring out \( bc \):
\[ bc(2 \cos A + 2 - k) = 0 \]
Since \( bc \neq 0 \) in a triangle, we can simplify to:
\[ 2 \cos A + 2 - k = 0 \]

### Step 6: Solve for \( \cos A \)
Rearranging gives:
\[ 2 \cos A = k - 2 \]
Thus:
\[ \cos A = \frac{k - 2}{2} \]

### Step 7: Determine the range of \( \cos A \)
Since \( \cos A \) must lie within the range \([-1, 1]\), we set up the inequalities:
\[ -1 \leq \frac{k - 2}{2} \leq 1 \]

### Step 8: Solve the inequalities
1. From the left inequality:
\[ -1 \leq \frac{k - 2}{2} \implies -2 \leq k - 2 \implies k \geq 0 \]
2. From the right inequality:
\[ \frac{k - 2}{2} \leq 1 \implies k - 2 \leq 2 \implies k \leq 4 \]

### Step 9: Combine the results
Combining both results, we find:
\[ 0 \leq k < 4 \]

### Conclusion
Thus, the final result is that \( k \) must satisfy:
\[ k \text{ is greater than or equal to } 0 \text{ and less than } 4. \]