# If the bisectors of the base angles of a triangle enclose an angle of $135^o$. Prove that the triangle is a right triangle.

Given:

The bisectors of the base angles of a triangle enclose an angle of $135^o$.

To do:

We have to prove that the triangle is a right triangle.

Solution:

Let in $\triangle ABC, OB$ and $OC$ be the bisectors of $\angle B$ and $\angle C$ and $\angle BOC = 135^o$

$\angle A+\angle B+\angle C=180^{\circ}$

Dividing both sides by 2, we get,

$\frac{1}{2} \angle A+\frac{1}{2} \angle B+\frac{1}{2} \angle C=180^{\circ}$

$\frac{1}{2} \angle A+\angle O B C+\angle O B C=90^{\circ}$

$\angle O B C+\angle O C B=90^{\circ}-\frac{1}{2}A$

In $\triangle B O C$,

$\angle B O C+\angle O B C+\angle O C B=180^{\circ}$

$\angle B O C+90^{\circ}-\frac{1}{2} \angle A=180^{\circ}$

$\angle B O C=90^{\circ}+\frac{1}{2} \angle A$

This implies,

$90^o+ \frac{1}{2}\angle A = 135^o$

$\frac{1}{2}\angle A= 135^o -90^o = 45^o$

Therefore,

$\angle A = 2(45^o) = 90^o$

Hence, $\triangle ABC$ is a right-angled triangle.

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Updated on: 10-Oct-2022

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