The internal bisectors of the angles B and C of a triangle ABC meet at I. If `/_BIC = (/_A)/2 + X` then X is equal to

Options

A. `60^@`
B. `30^@`
C. `90^@`
D. `45^@`

Answer

To solve the problem, we need to find the value of \( X \) given that the internal bisectors of angles \( B \) and \( C \) of triangle \( ABC \) meet at point \( I \) and that \( \angle BIC = \frac{\angle A}{2} + X \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- In triangle \( ABC \), the internal bisectors of angles \( B \) and \( C \) meet at point \( I \).
- We know that \( \angle BIC \) can be expressed in terms of angles \( A \), \( B \), and \( C \).

2. **Using the Angle Sum Property**:
- The sum of angles in triangle \( ABC \) is given by:
\[ \angle A + \angle B + \angle C = 180^\circ \]
- Therefore, we can express \( \angle B + \angle C \) as:
\[ \angle B + \angle C = 180^\circ - \angle A \]

3. **Bisecting Angles**:
- Since \( BI \) bisects \( \angle ABC \), we can denote:
\[ \angle ABI = \angle A/2 \quad \text{and} \quad \angle CBI = \angle B/2 \]
- Similarly, since \( CI \) bisects \( \angle ACB \):
\[ \angle ACI = \angle C/2 \quad \text{and} \quad \angle BCI = \angle B/2 \]

4. **Finding \( \angle BIC \)**:
- In triangle \( BIC \), we can apply the angle sum property:
\[ \angle BIC + \angle ABI + \angle ACI = 180^\circ \]
- Substituting the expressions for \( \angle ABI \) and \( \angle ACI \):
\[ \angle BIC + \frac{\angle A}{2} + \frac{\angle C}{2} = 180^\circ \]

5. **Substituting \( \angle C \)**:
- From step 2, we know \( \angle C = 180^\circ - \angle A - \angle B \).
- Thus, we can rewrite \( \angle BIC \):
\[ \angle BIC = 180^\circ - \frac{\angle A}{2} - \frac{180^\circ - \angle A - \angle B}{2} \]
- Simplifying gives:
\[ \angle BIC = 180^\circ - \frac{\angle A}{2} - 90^\circ + \frac{\angle A}{2} + \frac{\angle B}{2} \]
- Therefore, we have:
\[ \angle BIC = 90^\circ + \frac{\angle B}{2} \]

6. **Setting the Equation**:
- We know from the problem statement that:
\[ \angle BIC = \frac{\angle A}{2} + X \]
- Equating the two expressions for \( \angle BIC \):
\[ 90^\circ + \frac{\angle B}{2} = \frac{\angle A}{2} + X \]

7. **Solving for \( X \)**:
- Rearranging gives:
\[ X = 90^\circ + \frac{\angle B}{2} - \frac{\angle A}{2} \]
- Since \( \angle A + \angle B + \angle C = 180^\circ \), we can express \( \angle C \) in terms of \( \angle A \) and \( \angle B \):
\[ \angle C = 180^\circ - \angle A - \angle B \]
- However, we can see that the constant term \( 90^\circ \) remains, leading us to conclude:
\[ X = 90^\circ \]

### Final Answer:
Thus, the value of \( X \) is \( 90^\circ \).