Question

Given: △ABC,
Bisectors of ∠A and ∠B meet at O
Prove: m∠AOB>90°

Answer 1

The proof is given below.

Explanation:

Let the measure of angle A be
`2x` and B be
`2y`.

Now, from the triangle ΔABC,

`m\angle A+m\angle B+m\angle C=180\\2x+2y+m\angle C = 180\\2(x+y)=180-m\angle C`

As angle C is a positive quantity, so,
`2(x+y)` will be less than 180.

Therefore,
`2(x+y)<180\\x+y<(180)/(2)\\x+y<90`....... 1

Now, for triangle ΔAOB,

Since AO is an angle bisector of
`m\angle A`, therefore,
`m\angle OAB = (2x)/(2)=x`

Similarly,
`m\angle ABO = (2y)/(2)=y`

Now, sum of all the interior angles of a triangle is 180 degrees.

∴
`m\angle AOB + m\angle ABO + m\angle OAB=180`

`m\angle AOB +y+x=180\\m\angle AOB =180-(x+y)`

Now, consider the inequality 1.

`x+y<90`

Multiplying by -1 on both sides. This changes the sign of the inequality.

`-(x+y)>-90`

Adding 180 on both sides, we get

`180-(x+y)>180-90\\180-(x+y)>90`

But,
`180-(x+y)=m\angle AOB`

Therefore,
`m\angle AOB>90`. Hence, it is proved.