In triangle ABC if the line joining incenter to the circumcenter is parallel to the base BC, then the value of `(cosB + cosC-1)` is equal to

Options

A. 1
B. 2
C. 3
D. 0

Answer

To solve the problem, we need to analyze the given conditions in triangle ABC where the line joining the incenter (I) to the circumcenter (O) is parallel to the base BC. We are required to find the value of \( \cos B + \cos C - 1 \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let triangle ABC be given with incenter I and circumcenter O. The line joining I and O is parallel to the base BC.

2. **Using the Cosine Rule**:
- From the geometry, we can establish that in triangle OLB (where L is the foot of the perpendicular from O to BC), the cosine of angle A can be expressed as:
\[ \cos A = \frac{OL}{OB} \]
- Here, \( OL \) is the distance from the circumcenter O to the line BC, and \( OB \) is the circumradius R.

3. **Relating Inradius and Circumradius**:
- Since \( OL \) is parallel to BC, we can express \( OL \) in terms of the circumradius R:
\[ OL = R \cos A \]
- Therefore, we have:
\[ \cos A = \frac{R \cos A}{R} = \cos A \]

4. **Using the Inradius Formula**:
- The inradius \( r \) can be expressed using the formula:
\[ r = \frac{4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{a + b + c} \]
- However, we can also relate \( r \) to \( R \) using:
\[ r = R \cos A \]

5. **Equating the Two Expressions**:
- Setting the two expressions for \( r \) equal gives:
\[ R \cos A = \frac{4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{a + b + c} \]
- Cancelling \( R \) (assuming \( R \neq 0 \)):
\[ \cos A = \frac{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{a + b + c} \]

6. **Using the Identity**:
- We know from trigonometric identities that:
\[ \cos A + \cos B + \cos C - 1 = 0 \]
- This implies:
\[ \cos B + \cos C = 1 \]

7. **Finding the Required Value**:
- Now substituting back into our expression:
\[ \cos B + \cos C - 1 = 1 - 1 = 0 \]

### Final Answer:
Thus, the value of \( \cos B + \cos C - 1 \) is \( \boxed{0} \).