In a parallelogram, the bisectors of any two consecutive angles intersect at right angle.

Explanation

Let `ABCD` is a parallelogram
We know that
`∠A+∠B=180^@`
`OA` bisects `∠DAB` & `OB` bisects `∠CBA`
To prove: `∠AOB=90^@`
Proof: In `ΔAOB`,
`∠OAB+∠CBA+∠AOB=180^@`
`1/2∠DAB+1/2∠CBA+∠AOB=180^@`
`1/2(∠DAB+CBA)+∠AOB=180^@`
`1/2×180^@+∠AOB=180^@`
`90^@+∠AOB=180^@`
`∠AOB=180^@−90^@`
`∠AOB=90^@`
Hence proved.