Question

In a parallelogram ABCD, AO and BO are angle bisectors of angle A and angleB.if angle b>a,then prove that OA>OB

Answer 1

Step-by-step explanation:

Let a ∆ABC

Let the exterior angle C = 112 °

Now interior angle C= 68°

Sum of three angles of ∆ =180°_________(i)

Now , angle A: angle B= 3:4

So let angle A=3x

angle B= 4x

Now using (i)

3x + 4x + 68°= 180°

7x+ 68°= 180°

7x= 112°

x= 16

Now angles are A= 48°

B=64°

C=68°~

Answer 2

Answer:

AO<OB

Step-by-step explanation:

this is right solution...