Question

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles

Answer 1

See the given figure in picture

=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°

Hope it helps...... Please mark it as brainliest.....

Answer 2

Answer:

Answer:

we know that,

                      ABCD is a ||gm

Now, angle A + angle B =    $180^{0}$

        1/2 angle A + 1/2 angle B = $90^{0}$ --(1)

In Triangle AOB,

angle BAO + angle AOB + angle ABO = 180 --(Angle sum prop.)

We take 1/2 angle A instead of angle BAO

WE take 1/2 angle B instead of angle ABO

1/2 A + 1/2 B + angle AOB = 180 --(Angle sum prop.)

angle AOB = 180 - ( 1/2 A + 1/2 B ) -- from 1

angle AOB = 180 - 90

angle AOB = $90^{0}$

∴ Angle Bisecting of 2 adjacent angles intersect at $90^{0}$