Two parallel lines AB and CD are intersected by a transversal line EF at M and N, respectively. If the lines MP and NP are the bisectors of the interior angles BMN and DNM on the same side of the transversal, then `angleMPN` is equal to

Options

A. `180^(@)`
B. `60^(@)`
C. `90^(@)`
D. None of these

Answer

To solve the problem, we need to find the measure of angle \( MPN \) given that \( MP \) and \( NP \) are the angle bisectors of angles \( BMN \) and \( DNM \) respectively, formed by the intersection of the transversal \( EF \) with the parallel lines \( AB \) and \( CD \).

### Step-by-Step Solution:

1. **Identify the Angles**:
- Let \( \angle BMN = x \) and \( \angle DNM = y \).
- Since \( AB \) and \( CD \) are parallel lines and \( EF \) is a transversal, the interior angles on the same side of the transversal are supplementary:
\[ x + y = 180^\circ \quad \text{(1)} \]

2. **Find the Bisected Angles**:
- The angle bisector \( MP \) divides \( \angle BMN \) into two equal parts:
\[ \angle PMN = \frac{x}{2} \quad \text{(2)} \]
- The angle bisector \( NP \) divides \( \angle DNM \) into two equal parts:
\[ \angle PNM = \frac{y}{2} \quad \text{(3)} \]

3. **Substitute into the Supplementary Angles**:
- From equation (1), we can express \( y \):
\[ y = 180^\circ - x \]
- Substitute \( y \) into equation (3):
\[ \angle PNM = \frac{180^\circ - x}{2} = 90^\circ - \frac{x}{2} \quad \text{(4)} \]

4. **Sum of Angles in Triangle \( PMN \)**:
- In triangle \( PMN \), the sum of the angles is:
\[ \angle PMN + \angle PNM + \angle MPN = 180^\circ \quad \text{(5)} \]
- Substitute equations (2) and (4) into equation (5):
\[ \frac{x}{2} + \left(90^\circ - \frac{x}{2}\right) + \angle MPN = 180^\circ \]
- Simplifying this:
\[ 90^\circ + \angle MPN = 180^\circ \]
- Therefore:
\[ \angle MPN = 180^\circ - 90^\circ = 90^\circ \]

### Conclusion:
Thus, the measure of angle \( MPN \) is \( 90^\circ \).