The bisectors of two adjacent angles of a parallelogram intersect at

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 △=180]

⇒ 90° + ∠APB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)]

Hence, ∠APB = 90°