In an isosceles triangle $\mathrm{ABC}$, with $\mathrm{AB}=\mathrm{AC}$, the bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$ intersect each other at $O$. Join $A$ to $O$. Show that :(i) $\mathrm{OB}=\mathrm{OC}$(ii) $\mathrm{AO}$ bisects $\angle \mathrm{A}$

Given:

In an isosceles triangle  $ABC$, with $AB=A$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$.

To do:

We have to show that

(i) $OB=OC$

(ii) $AO$ bisects $\angle A$.

Solution:

(i) We know that,

In an isosceles triangle all the angle are equal.

This implies,

$\angle B= \angle C$

$\frac{1}{2}\angle B=\frac{1}{2}C$

This implies,

$\angle OBC=\angle OCB$

Therefore, since opposite side to the equal angles are equal we get,

$OB=OC$.

(ii) Let us consider $\triangle AOB$ and $\triangle AOC$,

We know,

From Side-Side-Side congruence rule states that if three sides of one triangle are equal to three corresponding sides of another triangle, then the triangles are congruent.

Given,

$AB=AC$ and we also have $OB=OC$

Since, $AO$ is the common side,

$AO=OA$

Therefore,

$\triangle AOB \cong \triangle AOC$

We also know,

From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding  sides must be equal.

Therefore,

$\angle BAO=\angle CAO$.

Therefore,

$AO$ bisects $\angle A$.

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Updated on: 10-Oct-2022

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