Let ABCD is a parallelogram.

AE and AD is the bisector angles of adjacent angles of ∠A and ∠D.

As we know that,

∠A + ∠D = 180^o (Sum of interior angles on the same side of traversal is 180^o)

∠A + ∠D =

= 90^o ---------------(i)

Now, in triangle AOD,

∠AED + ∠A + ∠D = 180^o (AE and AD is the angle bisector of ∠A and ∠D.)

∠AED + 90^o = 180^o (From eq (i))

∠AED = 180^o - 90^o = 90^o

So, the bisectors of any two adjacent angles of a parallelogram intersect at 90^o