In `DeltaABC,angleC=70^(@)` . The bisectors of `angleAandangleB` meet BC and AC in points D and E . Respectively .AD and BE intersect each other at P. What is the measure of `angleDPB`?

Options

A. `55^(@)`
B. `65^(@)`
C. `35^(@)`
D. `45^(@)`

Answer

To solve the problem, we need to find the measure of angle \( DPB \) in triangle \( ABC \) where \( \angle C = 70^\circ \). The angle bisectors of \( \angle A \) and \( \angle B \) meet \( BC \) and \( AC \) at points \( D \) and \( E \) respectively, and they intersect at point \( P \).

### Step-by-Step Solution:

1. **Identify the Angles of Triangle \( ABC \)**:
- Given \( \angle C = 70^\circ \).
- Let \( \angle A = a \) and \( \angle B = b \).
- By the triangle sum property, \( a + b + 70^\circ = 180^\circ \).
- Therefore, \( a + b = 110^\circ \).

2. **Use the Angle Bisector Theorem**:
- The angle bisector of \( \angle A \) divides it into two equal angles: \( \angle DAP = \frac{a}{2} \) and \( \angle PAD = \frac{a}{2} \).
- The angle bisector of \( \angle B \) divides it into two equal angles: \( \angle EBP = \frac{b}{2} \) and \( \angle PBE = \frac{b}{2} \).

3. **Calculate \( \angle APB \)**:
- The angle \( \angle APB \) can be expressed as:
\[ \angle APB = \angle DAP + \angle EBP = \frac{a}{2} + \frac{b}{2} = \frac{a + b}{2} = \frac{110^\circ}{2} = 55^\circ. \]

4. **Find \( \angle DPB \)**:
- Since \( P \) is the intersection of the angle bisectors, we can use the property that the angles around point \( P \) sum up to \( 360^\circ \).
- The angles around point \( P \) are \( \angle APB \), \( \angle DPA \), and \( \angle DPB \):
\[ \angle DPA = 180^\circ - \angle C = 180^\circ - 70^\circ = 110^\circ. \]
- Therefore, we can write:
\[ \angle DPB = 180^\circ - \angle APB - \angle DPA = 180^\circ - 55^\circ - 110^\circ. \]
- Simplifying this gives:
\[ \angle DPB = 180^\circ - 165^\circ = 15^\circ. \]

5. **Final Calculation**:
- However, we need to check the calculation again. We realize that we should have calculated:
\[ \angle DPB = 180^\circ - \angle APB = 180^\circ - 55^\circ = 125^\circ. \]
- Thus, the measure of \( \angle DPB \) is \( 125^\circ \).

### Conclusion:
The measure of \( \angle DPB \) is \( 55^\circ \).