Question

in a parallelogram prove that the angle bisector of two adjacent angle intersect

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°.
Hope it may help you!!!