Answer:
The proof is given below.
Step-by-step explanation:
Let the measure of angle A be [tex]2x[/tex] and B be [tex]2y[/tex].
Now, from the triangle ΔABC,
[tex]m\angle A+m\angle B+m\angle C=180\\2x+2y+m\angle C = 180\\2(x+y)=180-m\angle C[/tex]
As angle C is a positive quantity, so, [tex]2(x+y)[/tex] will be less than 180.
Therefore, [tex]2(x+y)<180\\x+y<\frac{180}{2}\\x+y<90[/tex]....... 1
Now, for triangle ΔAOB,
Since AO is an angle bisector of [tex]m\angle A[/tex], therefore, [tex]m\angle OAB = \frac{2x}{2}=x[/tex]
Similarly, [tex]m\angle ABO = \frac{2y}{2}=y[/tex]
Now, sum of all the interior angles of a triangle is 180 degrees.
∴ [tex]m\angle AOB + m\angle ABO + m\angle OAB=180[/tex]
[tex]m\angle AOB +y+x=180\\m\angle AOB =180-(x+y)[/tex]
Now, consider the inequality 1.
[tex]x+y<90[/tex]
Multiplying by -1 on both sides. This changes the sign of the inequality.
[tex]-(x+y)>-90[/tex]
Adding 180 on both sides, we get
[tex]180-(x+y)>180-90\\180-(x+y)>90[/tex]
But, [tex]180-(x+y)=m\angle AOB[/tex]
Therefore, [tex]m\angle AOB>90[/tex]. Hence, it is proved.
