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