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 $\angle A$.

Updated on: 10-Oct-2022

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