Prove that the bisectors of any two adjacent angles of a parallelogram are at right angle.

Given: ABCD is a parallelogram. AO is the angle bisector of ∠DAB and BO is the angle bisector of ∠ABC.

To prove: ∠AOB = 90°

Proof: AB ∥ CD and AD ∥ BC

Sum of corresponding interior angles is 180°.

⇒ ∠DAB + ∠ABC = 180°

⇒ $\dfrac{1}{2}$ ∠DAB + $\dfrac{1}{2}$ ∠ABC = $\dfrac{1}{2}$ x 180°

⇒ ∠OAB + ∠OBA = 90° ………..(1)

In Δ AOB, sum of all angles in triangle is 180°.

⇒ ∠OAB + ∠OBA + ∠BOA = 180°

⇒ 90° + ∠BOA = 180°

⇒ ∠BOA = 180° - 90°

⇒ ∠BOA = 90°

Hence, the bisectors of any two adjacent angles of a parallelogram are at right angle.