The external bisector of `/_B and /_C` of `DeltaABC `(where AB and AC extended to E and F respectively) meet at point P. If `/_BAC = 100^@`, then the measure of `/_BPC` is

Options

A. `50^@`
B. `80^@`
C. `40^@`
D. `100^@`

Answer

To solve the problem, we need to find the measure of angle \( \angle BPC \) given that \( \angle BAC = 100^\circ \). We will use the property of external angle bisectors.

### Step-by-step Solution:

1. **Identify the Given Information**:
- We have triangle \( ABC \).
- The external bisectors of angles \( B \) and \( C \) meet at point \( P \).
- The measure of angle \( A \) (i.e., \( \angle BAC \)) is given as \( 100^\circ \).

2. **Use the External Angle Bisector Theorem**:
- The theorem states that the measure of the angle formed by the external bisectors of angles \( B \) and \( C \) at point \( P \) can be calculated using the formula:
\[ \angle BPC = 90^\circ - \frac{\angle A}{2} \]

3. **Substitute the Given Angle**:
- We know \( \angle A = 100^\circ \).
- Therefore, we substitute \( \angle A \) into the formula:
\[ \angle BPC = 90^\circ - \frac{100^\circ}{2} \]

4. **Calculate \( \frac{100^\circ}{2} \)**:
- This simplifies to:
\[ \frac{100^\circ}{2} = 50^\circ \]

5. **Final Calculation**:
- Now substitute \( 50^\circ \) back into the equation:
\[ \angle BPC = 90^\circ - 50^\circ = 40^\circ \]

6. **Conclusion**:
- Thus, the measure of \( \angle BPC \) is \( 40^\circ \).

### Final Answer:
\[ \angle BPC = 40^\circ \]