Question

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

Answer 1

Answer:

The proof is given below.

Step-by-step 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, $2(x+y)$....... 1

Now, for triangle ΔAOB,

Since AO is an angle bisector of $m\angle A$, therefore, $m\angle OAB = \frac{2x}{2}=x$

Similarly, $m\angle ABO = \frac{2y}{2}=y$

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.

$x+y$

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.