In a `DeltaABC, (a + b + c) (b + c - a) = lambda bc`. (where symbols have their usual meaning)& `lambda in I`, then greatest value of `lambda` is

Options

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

Answer

To solve the problem step by step, we start with the given equation for triangle \( ABC \):

\[ (a + b + c)(b + c - a) = \lambda bc \]

### Step 1: Expand the left-hand side
We can expand the left-hand side of the equation:

\[ (a + b + c)(b + c - a) = (b + c)^2 - a^2 \]

### Step 2: Rewrite the equation
Now, we can rewrite the equation as:

\[ (b + c)^2 - a^2 = \lambda bc \]

### Step 3: Apply the identity
Using the identity \( (b + c)^2 = b^2 + c^2 + 2bc \), we substitute into the equation:

\[ b^2 + c^2 + 2bc - a^2 = \lambda bc \]

### Step 4: Rearranging the equation
Rearranging gives us:

\[ b^2 + c^2 - a^2 + 2bc = \lambda bc \]

### Step 5: Isolate \( \lambda \)
Now, isolate \( \lambda \):

\[ \lambda = \frac{b^2 + c^2 - a^2 + 2bc}{bc} \]

### Step 6: Simplify the expression
This can be simplified further:

\[ \lambda = \frac{b^2 + c^2 + 2bc - a^2}{bc} \]

### Step 7: Use the cosine rule
Using the cosine rule, we know that:

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

Thus, we can express \( \lambda \) in terms of \( \cos A \):

\[ \lambda = 2 \cos A + 2 \]

### Step 8: Determine the maximum value of \( \lambda \)
The maximum value of \( \cos A \) is 1 (when \( A = 0^\circ \)). Therefore, the maximum value of \( \lambda \) is:

\[ \lambda_{\text{max}} = 2 \cdot 1 + 2 = 4 \]

### Conclusion
Thus, the greatest value of \( \lambda \) is:

\[ \boxed{4} \]

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