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# 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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