Question

in a parallelogram ABCD angle bisector of angle A and Angle B intersect at O prove that angle aob is equal to 90 degree​

Answer 1

Given : in a parallelogram ABCD angle bisector of ∠A and ∠B intersect at O

To Find : prove that ∠AOB is equal to 90°

Solution:

ABCD is a parallelogram

Sum of adjacent angles of a parallelogram = 180°

∠A + ∠B  = 180°

Dividing both sides by 2

=> ∠A/2 + ∠B/2 = 90°

Bisector of ∠A    = ∠BAO = ∠A/2

Bisector of ∠B    = ∠ABO = ∠B/2

∠BAO +  ∠ABO  + ∠AOB = 180°    ( Sum of angles of a triangle)

=> ∠A/2 +  ∠B/2  + ∠AOB = 180°

=> 90° + ∠AOB = 180°    Substitute ∠A/2 + ∠B/2 = 90°

=>  ∠AOB =90°

QED

Hence Proved

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Answer 2

Step-by-step explanation:

whoever has answered the question thank you