Question

In a parallelogram the sum of the angle bisectors of two adjacent angle is:

Answer 1

Given: ABCD is a llgm.

Ao is the bisector of angle A

BO is the bisector of angle B

To prove : angle o= 90°

Proof: Angle A+ Angle B=180°(co interior angles )

Now,

1/2A+1/2B =90°

Now in triangle AOB,

angle ABO + angle BAO + angle OAB = 180° ( ANGLE SUM PROPERTY )

Angle OAB= 180°- (angle BAO + angle ABO )

Angle OAB= 180° -(90°) [above proved]

Angle OAB= 90°

HENCE, PROVED