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 $\angle A$
$(i)$ $OB = OC$
$(ii)$ $AO$ bisects $\angle A$
(N/A) $(i)$ In $\Delta ABC$,we have
$AB = AC$ [Given]
$\therefore \angle C = \angle B$ [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 $\frac{1}{2} \angle B = \frac{1}{2} \angle C$]
$\therefore$ Using $SAS$ congruence criteria,
$\Delta ABO \cong \Delta ACO$
$\Rightarrow \angle OAB = \angle OAC$ [c.p.c.t.]
$\Rightarrow AO$ bisects $\angle A$.
$AB = AC$ [Given]
$\therefore \angle C = \angle B$ [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 $\frac{1}{2} \angle B = \frac{1}{2} \angle C$]
$\therefore$ Using $SAS$ congruence criteria,
$\Delta ABO \cong \Delta ACO$
$\Rightarrow \angle OAB = \angle OAC$ [c.p.c.t.]
$\Rightarrow AO$ bisects $\angle A$.
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