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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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