In an isosceles triangle $ A B C $, with $ A B=A C $, the bisectors of $ \angle B $ and $ \angle C $ intersect each other at $ O . $ Join $ A $ to $ O $. Show that :
(1) $ \mathrm{OB}=\mathrm{OC} $
(ii) AO bisects $ \angle \mathrm{A} $
GIVEN: In an isosceles triangle \( A B C \), with \( A B=A C \), the bisectors of \( \angle B \) and \( \angle C \) intersect each other at \( O . \)
To Do:
(i) \( \mathrm{OB}=\mathrm{OC} \)
(ii) AO bisects \( \angle \mathrm{A} \)
Solution:
Given that AB = AC, ABC is an isosceles triangle
In $\Delta$ ABC, AB = AC. So opposite angles to these equal sides are also equal
i.e., $\angle$B = $\angle$C
As OB and OC are bisectors of these angles B and C, $\angle$OBC = $\angle$OCB as
$\angle\frac{B}{2}$ = $\angle\frac{C}{2}$
(i)
In $\Delta$ OBC, $\angle$OBC = $\angle$OCB
$\Rightarrow$ OB = OC as sides opposite to equal angles must also be equal. Hence proved
(ii) To prove AO bisects $\angle$A
The point of intersection of angle bisectors of angles B and C is O and it is the incenter of the triangle and the third angle bisector passes through O. AO passes through O. So AO bisects $\angle$A. Hence proved.
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