Math, asked by montikansal8633, 1 year ago

Prove that in a parallelogram the bisectors of consecutive angles form a right angle

Answers

Answered by naiyarrajesh
0

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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