ABCD is a parallelogram.The bisectors of angle A and angle B meet at O. prove that angle AOB Is 90 degrees

Dear Student!

Here is the answer to your query.

Given : ABCD is a parallelogram in which AO and BO are angle bisectors of ∠A and ∠B

Now since ABCD is a parallelogram

∴ AD||BC

Now AD||BC and transversal AB intersect them

∴∠ A + ∠B = 180° (∴ sum of consecutive interior angle is 180° )

⇒ ∠1 + ∠2 = 90°  (∴ AO and BO are angle bisectors)  ... (1)

In ΔAOB we have

∠1 + ∠AOB + ∠2 = 180°

⇒90°  +∠AOB = 180°  (from (1))

⇒ ∠AOB = 180° – 90° = 90°

Cheers!