Question

in fig AO,BO bisects angle A and angle B of the parallelogram ABCD. prove that angle aob equal to 90 degree​

Answer 1

Answer:

ABCD is a quadrilateral

To prove : ∠AOB= 90 degree

AO and BO is bisector of A and B

∠1=∠2∠3=∠4...(1)

∠A+∠B+∠C+∠D=360

(Angle sum property)

1/2 (∠A+∠B+∠C+∠D)=180...(2)

In △AOB

∠1+∠3+∠5= 1/2 (∠A+∠B+∠C+∠D)

∠1+∠3+∠5=∠1+∠3+ 1/2(∠C+∠D)

∠AOB= 1/2 (∠C+∠D)

hence proved