Two parallel lines AB and CD are intersected by a transversal EF at M and N respectively.
The lines MP and NP are the bisectors of interior angles $∠ B M N$ and $∠ D N M$ on the same side of the transversal. Then $∠ M P N$ is equal to:

60^{\circ}

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90^{\circ}

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45^{\circ}

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120^{\circ}

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Solution

The correct option is B $90^{\circ}$
Let $∠ B M N = 2 x$ and $∠ D N M = 2 y$
$2 x + 2 y = 180^{\circ}$

$\Rightarrow x + y = 90^{\circ}$
$∴ ∠ M P N = 180^{\circ} - 90^{\circ} = 90^{\circ}$

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Two parallel lines AB and CD are intersected by a transversal EF at M and N respectively. The lines MP and NP are the bisectors of interior angles ∠BMN and ∠DNM on the same side of the transversal. Then ∠MPN is equal to:

The correct option is B 90∘Let ∠BMN=2x and ∠DNM=2y 2x+2y=180∘ ⇒x+y=90∘ ∴∠MPN=180∘−90∘=90∘

Two parallel lines AB and CD are intersected by a transversal EF at M and N respectively.
The lines MP and NP are the bisectors of interior angles $∠ B M N$ and $∠ D N M$ on the same side of the transversal. Then $∠ M P N$ is equal to:

The correct option is B $90^{\circ}$
Let $∠ B M N = 2 x$ and $∠ D N M = 2 y$
$2 x + 2 y = 180^{\circ}$

$\Rightarrow x + y = 90^{\circ}$
$∴ ∠ M P N = 180^{\circ} - 90^{\circ} = 90^{\circ}$