`Delta ABC` ,angle B external angle C meets at P `angle A = 80 " then " angle BPC ` = ?

Answer

To solve the problem, we need to find the measure of angle BPC in triangle ABC, given that angle A is 80 degrees and we are dealing with the external bisectors of angles B and C.

### Step-by-Step Solution:

1. **Understanding the Triangle and Given Information**:
- We have triangle ABC with angle A = 80 degrees.
- We need to find angle BPC where P is the intersection of the external bisectors of angles B and C.

2. **Finding the Sum of Angles in Triangle ABC**:
- The sum of the angles in any triangle is 180 degrees.
- Therefore, we can write:
\[ \text{Angle A} + \text{Angle B} + \text{Angle C} = 180^\circ \]
- Substituting the known value:
\[ 80^\circ + \text{Angle B} + \text{Angle C} = 180^\circ \]
- This simplifies to:
\[ \text{Angle B} + \text{Angle C} = 180^\circ - 80^\circ = 100^\circ \]

3. **Finding the External Angles**:
- The external angle at B (let's denote it as angle B') and the external angle at C (angle C') can be calculated as:
\[ \text{Angle B'} = 180^\circ - \text{Angle B} \]
\[ \text{Angle C'} = 180^\circ - \text{Angle C} \]
- Since we know that:
\[ \text{Angle B} + \text{Angle C} = 100^\circ \]
- Therefore:
\[ \text{Angle B'} + \text{Angle C'} = (180^\circ - \text{Angle B}) + (180^\circ - \text{Angle C}) = 360^\circ - 100^\circ = 260^\circ \]

4. **Finding the Measure of Angles B' and C'**:
- Since angles B' and C' are external angles, and they are bisected at point P:
\[ \text{Angle B'} + \text{Angle C'} = 260^\circ \]
- If we let angle B' = x and angle C' = y, then:
\[ x + y = 260^\circ \]
- Since P is the intersection of the bisectors:
\[ \text{Angle BPC} = 180^\circ - \frac{x + y}{2} \]
- Substituting the value of x + y:
\[ \text{Angle BPC} = 180^\circ - \frac{260^\circ}{2} = 180^\circ - 130^\circ = 50^\circ \]

5. **Final Result**:
- Therefore, the measure of angle BPC is:
\[ \text{Angle BPC} = 50^\circ \]

### Summary:
The measure of angle BPC is 50 degrees.