Question

In parallelogram Show that the angle bisector of two adjacent angles intersect at right angles
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Answer 1

Step-by-step explanation:

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°

Answer 2

Answer:

extend the line ae to c and the line be to d

since ac and bd are the diagonals of the parallelogram they bisect each other

so the angle formed between them is a right angle therefore ae and be form right angle

hope it was helpful