Question

Prove that in a parallelogram the angle bisectors of two adjacent angles intersect at right angle.

Answer 1

It is given that ABCD is a parallelogram.
∴ $\angle$DAB + $\angle$CBA = 180$°$ (Angles on the same side of a transversal line are supplementary.)
Dividing both sides by 2, we get:
$\angle$OAB + $\angle$ABO = 90$°$ (OA and OB are bisectors of $\angle$DAB and $\angle$CBA)

Now, consider $∆$AOB
$\angle$OAB + $\angle$ABO + $\angle$BOA = 180$°$ [Angle sum property]
or,
90$°$ + $\angle$AOB = 180$°$
or,
$\angle$AOB = 90$°$