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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Similar Questions
In $\Delta ABC$,$AD$ is the perpendicular bisector of $BC$ (see figure). Show that $\Delta ABC$ is an isosceles triangle in which $AB = AC$.
Medium
View Solution$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
Difficult
View Solution$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ bisects $\angle A$.
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ bisects $\angle A$.
Medium
View SolutionTwo sides $AB$ and $BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\Delta PQR$ (see Fig). Show that:
$(i)$ $\Delta ABM \cong \Delta PQN$
$(ii)$ $\Delta ABC \cong \Delta PQR$
$(i)$ $\Delta ABM \cong \Delta PQN$
$(ii)$ $\Delta ABC \cong \Delta PQR$
Medium
View SolutionIn the figure,$\angle B < \angle A$ and $\angle C < \angle D$. Show that $AD < BC$.
Difficult
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