Question

in the adjoining figure, ABCD is a parallelgram, AO and BO are the bisector of angle A and angle B respectively. Prove that LAOB =90 degree

Answer 1

Answer: as we no. Angle sum property of triangle is 180° and the angle of rectangular 90 degree each so the bisector will bisect each angle 45 and 45° I didn't triangle aob using angle sum property we can show that a 45 + 45 is equals to 90 degree

and then 180 - 90 equals to 90° angle aob is equals to 90 degree

Step-by-step explanation:

Answer 2

Answer:

∠AOB = 90° is answer for sure.

Step-by-step explanation:

As the opposite sides of parallelogram are complementary...

∠A + ∠B = 180°....................(1)

Seg AO is the angle bisector of ∠A.

∴∠OAB = 1/2 ∠ A....................(2)

Seg OB is angle bisector of ∠B.

∴∠OBA = 1/2 ∠B................(3)

In Δ AOB,

∠OAB +∠OBA+∠AOB= 180°........ angle sum property.

from (1),(2)&(3)

∠OAB + ∠OBA = 90°

∴ 90° + ∠ AOB = 180°

∴∠AOB= 90°

Hence proved.

hope i explained it well.