In a triangle ABC , the line joining the circumcentre and incentre is parallel to BC, then Cos B + Cos C is equal to:

Options

A. `3/2`
B. 1
C. `3/4`
D. `1/2`

Answer

To solve the problem, we need to find the value of \( \cos B + \cos C \) given that the line joining the circumcenter (O) and the incenter (I) of triangle ABC is parallel to side BC.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
Since the line joining the circumcenter (O) and the incenter (I) is parallel to side BC, we can use the relationship between the circumradius (R) and the inradius (r) in the triangle.

2. **Using the Cosine Rule**:
We know that in any triangle, the cosine of angle A can be expressed in terms of the circumradius and inradius:
\[ \cos A = \frac{r}{R} \]
where \( r \) is the inradius and \( R \) is the circumradius.

3. **Expressing the Inradius**:
We can express the inradius \( r \) in terms of the angles of the triangle:
\[ r = \frac{4R \cdot \sin A}{\sin B \cdot \sin C} \]

4. **Setting Up the Equation**:
From the condition that \( OI \parallel BC \), we can derive that:
\[ \frac{4 \sin A}{\sin B \cdot \sin C} = \cos A \]

5. **Using the Known Identity**:
We know a standard result in triangle geometry:
\[ 4 \cdot \frac{\sin A}{\sin B \cdot \sin C} = \cos A + \cos B + \cos C - 1 \]
Therefore, we can set this equal to \( \cos A \):
\[ \cos A + 1 = \cos A + \cos B + \cos C \]

6. **Cancelling \( \cos A \)**:
By rearranging the equation, we can cancel \( \cos A \) from both sides:
\[ 1 = \cos B + \cos C \]

7. **Conclusion**:
Thus, we find that:
\[ \cos B + \cos C = 1 \]

### Final Answer:
\[ \cos B + \cos C = 1 \]