Question

prove that bisector of any two adjacent angles of a parallelogram are at right angles

Answer 1

ABCD is a parallelogram. The angle bisectors AE and BE of adjacent angles A and B meet at E.



To Prove: ∠AEB = 90°
Proof: ∵ AD || BC
| Opposite sides of ||gm and transversal AB intersects them
∴ ∠DAB + ∠CBA = 180°
| ∵ Sum of consecutive interior angles on the same side of a transversal is 180°
⇒ 2∠EAB + 2∠EBA = 180°
| ∵ AE and BE are the bisectors of ∠DAB and ∠CBA respectively.
⇒ ∠EAB + ∠EBA = 90° ...(1)
In ∆EAB,
∠EAB + ∠EBA + ∠AEB = 180°
| ∵ The sum of the three angles of a triangle is 180°
⇒ 90° + ∠AEB = 180° | From (1)
⇒ ∠AEB = 90°.

Answer 2

Step-by-step explanation:

Let ABCD be a parallelogram where OA and OD are the bisectors of adjacent angles A &D

since ABCD is a parallelogram, AB parallel CD [ opposite sides of a parallelogram are parallel]

Also , angle BAD + angle CDA = 180 degree ( sum of the interior angles on the same sides of the transversal is 180 degree)

=> 1 /2 angle BAD + 1/2 angle CDA=1/2 × 180 degree

=>angle 1 +angle 2 = 90 degree (AO and DO are angle bisectors of angle A and angle D ) -1

In angle AQD ,

angle 1 + angle AOD + angle 2 =180 degree

=> angle AOD+ 90 degree= 180 degree

=>Angle AOD= 180-90

=>angle AOD = 90

In a parallelogram the bisectors of the adjacent angles intersect at right angle