Question

in parallelogram ABCD bisector of angle A and B intersect each other at O. prove that angle AOB is equal to 90 degree​

Answer 1

Apply Angle Sum property of Triangles

Step-by-step explanation:

Since ABCD is a parallelogram  

∴ AD  | | BC  

⇒ ∠A + ∠B = 180° [sum of consecutive interior angle]  

⇒ $\frac {1}{2}$∠A  + $\frac {1}{2}$∠B = 90°    

⇒ ∠1 + ∠2 = 90°  ---- (i)

[∵ AO is the bisector of ∠A and BO is the bisector of ∠B ]

∴ ∠1 = $\frac {1}{2}$∠A and ∠2 = $\frac {1}{2}$∠B]

Now, △AOB, we have  

∠1  + ∠AOB + ∠2 = 180° [sum of three angles of a triangle]

⇒  90° + ∠AOB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)]

Hence,   ∠AOB  =  90°  

Answer 2

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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