Question

The bisector of two adjacent angles of a parallelogram intersect at which angle ?

☘Don't spam☘
‼Need answer with explanation ‼✔✔
​

Answer 1

$\huge\star\mathfrak\red{{Hey Mate...!!}}$

● The bisector of two adjacent angles of a parallelogram intersect at right angle (i.e 90°).

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

Let the ABCD be parallelogram

Given:-

• ∠EAB = ½∠A

• ∠EBA = ½∠B

To Find:-

• Measure of ∠AEB

Solution:-

➥ As we know, Sum of adjacent angles of a parallelogram is 180°.

Hence,

∠ A + ∠B = 180°

➟ ½∠A + ½∠B = ½ of 180°

➟ ½∠A + ½∠B = 90°

➟ ∠EAB + ∠EBA = 90° ------- eq. 1

➥ As we know, Sum of all angles of a triangle is 180°

Hence,

Sum of angles of ∆ EAB = 180°

➟ ∠EAB + ∠EBA + ∠AEB = 180°

➟ ( ∠EAB + ∠EBA ) + ∠AEB = 180°

➟ 90° + ∠AEB = 180° [ By equation 1 ]

➟ ∠AEB = 180° - 90°

∴ ∠AEB = 90°

Answer:-

The bisector of two adjacent angles of a parallelogram intersect at right angle ( 90° ).

═══◄••❀••►═══◄••❀••►═══