If 2 parallel lines are cut by a transversal then prove that the interior angle bisector on same side of transversal intersect each other at right angle
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let the 2 interior angles be 2x and 2y
then, as lines are parallel
2x+2y=180°
2(x+y)=180°
(x+y)=180°/2
x+y= 90°
Hence Proved
then, as lines are parallel
2x+2y=180°
2(x+y)=180°
(x+y)=180°/2
x+y= 90°
Hence Proved
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Solutions:
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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