The bisectors of any two adjacent angles intersect at 90∘
Let ABCD be the parallelogram
OA and OB are the bisectors of ∠A and ∠B
From the figure, ∠A+∠B=180o [co-interior angles]
12∠A+12∠B=90o
From △AOB
∠ABO+∠BAO+∠OAB=180o [angle sum property]
∠OAB=180o−(∠BAO+∠ABO)
∠OAB=180o−90o [above proved]
⇒∠OAB=90o