Prove that `bc cos^(2).(A)/(2) + ca cos^(2).(B)/(2) + ab cos^(2).(C)/(2) = s^(2)`

Explanation

To prove that

\[ \frac{bc \cos^2 A}{2} + \frac{ca \cos^2 B}{2} + \frac{ab \cos^2 C}{2} = s^2, \]

where \(s\) is the semi-perimeter of the triangle, we will follow these steps:

### Step 1: Write the expressions for \(\cos^2 A\), \(\cos^2 B\), and \(\cos^2 C\)

Using the half-angle formulas, we can express \(\cos^2 A\), \(\cos^2 B\), and \(\cos^2 C\) in terms of the semi-perimeter \(s\) and the sides of the triangle:

\[ \cos^2 A = \frac{s(s-a)}{bc}, \quad \cos^2 B = \frac{s(s-b)}{ca}, \quad \cos^2 C = \frac{s(s-c)}{ab}. \]

### Step 2: Substitute these expressions into the left-hand side

Now we substitute these expressions into the left-hand side of the equation:

\[ \frac{bc \cdot \frac{s(s-a)}{bc}}{2} + \frac{ca \cdot \frac{s(s-b)}{ca}}{2} + \frac{ab \cdot \frac{s(s-c)}{ab}}{2}. \]

This simplifies to:

\[ \frac{s(s-a)}{2} + \frac{s(s-b)}{2} + \frac{s(s-c)}{2}. \]

### Step 3: Factor out \(\frac{s}{2}\)

Now we can factor out \(\frac{s}{2}\):

\[ \frac{s}{2} \left( (s-a) + (s-b) + (s-c) \right). \]

### Step 4: Simplify the expression inside the parentheses

Now, simplify the expression inside the parentheses:

\[ (s-a) + (s-b) + (s-c) = 3s - (a + b + c). \]

Since \(s = \frac{a+b+c}{2}\), we have:

\[ a + b + c = 2s. \]

Thus,

\[ 3s - (a + b + c) = 3s - 2s = s. \]

### Step 5: Substitute back into the equation

Now substituting back, we have:

\[ \frac{s}{2} \cdot s = \frac{s^2}{2}. \]

### Step 6: Final expression

Thus, we have:

\[ \frac{s^2}{2} = s^2. \]

This shows that the left-hand side equals the right-hand side, proving the equation.

### Conclusion

Therefore, we have proved that:

\[ \frac{bc \cos^2 A}{2} + \frac{ca \cos^2 B}{2} + \frac{ab \cos^2 C}{2} = s^2. \]