Prove that when a transversal intersects two parallel lines the bisectors of the interior angle on the same side of the transversal form a right angle at their point of intersection.
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sry I don't know. .......................
Mayyu2002:
Can anyone answer please?
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We know that the sum of interior angles on the same side of the transversal is 180°.
Hence, ∠BMN + ∠DNM = 180°
=> 1/2∠BMN + 1/2∠DNM = 90°
=> ∠PMN + ∠PNM = 90°
=> ∠1 + ∠2 = 90° ............. (i)
In △PMN, we have
∠1 + ∠2 + ∠3 = 180° ......... (ii)
From (i) and (ii), we have
90° + ∠3 = 180°
=> ∠3 = 90°
=> PM and PN intersect at right angles.
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