In `DeltaABC`, if the angles A.B.C are in AP, which one of the following is correct ?

Options

A. c=a+b
B. `c^(2)=a^(2)+b^(2)-ab`
C. `a^(2)=b^(2)+c^(2)-bc`
D. `b^(2)=a^(2)+c^(2)-ac`

Answer

To solve the problem, we need to analyze the angles of triangle ABC, given that angles A, B, and C are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Understanding Angles in AP**:
If angles A, B, and C are in AP, we can express them as:
- A = a
- B = a + d
- C = a + 2d
Here, 'a' is the first angle and 'd' is the common difference.

**Hint**: Recall that in an AP, the middle term is the average of the other two terms.

2. **Using the Triangle Sum Property**:
The sum of the angles in a triangle is always 180 degrees. Therefore, we can write:
\[ A + B + C = 180^\circ \]
Substituting the expressions for A, B, and C:
\[ a + (a + d) + (a + 2d) = 180^\circ \]
Simplifying this gives:
\[ 3a + 3d = 180^\circ \]

**Hint**: Factor out the common term to simplify the equation.

3. **Solving for a + d**:
Dividing the entire equation by 3:
\[ a + d = 60^\circ \]
This means that angle B, which is \(a + d\), is equal to 60 degrees.

**Hint**: Identify which angle corresponds to the expression you simplified.

4. **Using the Cosine Rule**:
We can now apply the cosine rule, which states:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Substituting B = 60 degrees:
\[ \cos 60^\circ = \frac{a^2 + c^2 - b^2}{2ac} \]
Since \(\cos 60^\circ = \frac{1}{2}\), we have:
\[ \frac{1}{2} = \frac{a^2 + c^2 - b^2}{2ac} \]

**Hint**: Remember to substitute the cosine value correctly.

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 1 \cdot 2ac = a^2 + c^2 - b^2 \]
Simplifying this results in:
\[ b^2 = a^2 + c^2 - ac \]

**Hint**: Keep track of the signs when rearranging the equation.

6. **Conclusion**:
Thus, we have derived that:
\[ b^2 = a^2 + c^2 - ac \]
Therefore, the correct option is:
- Option D: \(b^2 = a^2 + c^2 - ac\)

**Hint**: Verify your final equation matches one of the provided options.