In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th
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1 Answer

Answer :

Given : A parallelogram ABCD such that the bisectors of adjacent angles A and B intersect at P.  To prove : ∠APB = 90°   Proof : Since ABCD is a | | gm  ∴ AD  | | BC  ⇒ ∠A + ∠B = 180° [sum of consecutive interior angle]  ⇒ 1 / 2  ∠A  + 1 / 2  ∠B = 90°    ⇒ ∠1 + ∠2 = 90°  ---- (i) [∵ AP is the bisector of  ∠A and BP is the bisector of ∠B ] ∴ ∠1 = 1 / 2 ∠A and ∠2 = 1 / 2   ∠B] Now, △APB , we have  ∠1  + ∠APB + ∠2 = 180°  [sum of three angles of a △] ⇒   90° + ∠APB + ∠2 = 180° [ ∵  ∠1 + ∠2 =  90° from (i)] Hence,    ∠APB  =  90°   
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