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(N/A) $(i)$ In $\Delta ABC$,we have$AB = AC$ [Given]$	herefore \angle C = \angle B$ [Angles opposite to equal sides are equal]$\Rightarrow \frac{1}{2

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(N/A) $(i)$ In $\Delta ABC$,we have$AB = AC$ [Given]$	herefore \angle C = \angle B$ [Angles opposite to equal sides are equal]$\Rightarrow \frac{1}{2} \angle C = \frac{1}{2} \angle B$Or $\angle OCB = \angle OBC$$\Rightarrow OB = OC$ [Sides opposite to equal angles are equal]$(ii)$ In $\Delta ABO$ and $\Delta ACO$,we have$AB = AC$ [Given]$OB = OC$ [Proved above]$\angle OBA = \angle OCA$ [Since $\frac{1}{2} \angle B = \frac{1}{2} \angle C$]$	herefore$ Using $SAS$ congruence criteria,$\Delta ABO \cong \Delta ACO$$\Rightarrow \angle OAB = \angle OAC$ [c.p.c.t.]$\Rightarrow AO$ bisects $\angle A$.

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

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