In `triangleABC, (b+c-a)tan((A)/(2))=`

Options

A. `(A(triangleABC))/(s)`
B. `(2A(triangleABC))/(s)`
C. `A(triangleABC)`
D. `2A(triangleABC)`

Answer

To solve the problem \( (b + c - a) \tan\left(\frac{A}{2}\right) \) in triangle \( ABC \), we can follow these steps:

### Step 1: Understand the semi-perimeter
First, we need to define the semi-perimeter \( s \) of triangle \( ABC \):
\[ s = \frac{a + b + c}{2} \]

### Step 2: Express \( b + c - a \)
We can rewrite \( b + c - a \) in terms of the semi-perimeter:
\[ b + c - a = (a + b + c) - 2a = 2s - 2a = 2(s - a) \]

### Step 3: Use the formula for \( \tan\left(\frac{A}{2}\right) \)
The formula for \( \tan\left(\frac{A}{2}\right) \) is given by:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} \]

### Step 4: Substitute into the expression
Now substitute \( b + c - a \) and \( \tan\left(\frac{A}{2}\right) \) into the original expression:
\[ (b + c - a) \tan\left(\frac{A}{2}\right) = 2(s - a) \cdot \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} \]

### Step 5: Simplify the expression
We can simplify this expression:
\[ = 2 \sqrt{(s - b)(s - c)} \cdot \frac{s - a}{\sqrt{s(s - a)}} \]
The \( s - a \) terms will cancel out:
\[ = 2 \cdot \sqrt{(s - b)(s - c)} \cdot \frac{1}{\sqrt{s}} \]

### Step 6: Recognize the area of the triangle
Using Heron's formula, the area \( K \) of triangle \( ABC \) can be expressed as:
\[ K = \sqrt{s(s - a)(s - b)(s - c)} \]
Thus, we can express \( (s - b)(s - c) \) in terms of the area:
\[ (s - b)(s - c) = \frac{K^2}{s(s - a)} \]

### Final Expression
Substituting back, we find:
\[ (b + c - a) \tan\left(\frac{A}{2}\right) = 2 \cdot \frac{K}{s} \]

### Conclusion
Thus, the final result is:
\[ (b + c - a) \tan\left(\frac{A}{2}\right) = \frac{2K}{s} \]