Question

In a triangle ABC, AB = AC and the bisectors of angles B and C intersect at O. Prove that BO = CO and AO is the bisector of angle $\angle BAC$.

Answer 1

Since the angles opposite to equal sides are equal,

$\therefore AB=AC$

$\Rightarrow \angle C=\angle B$

$\Rightarrow \frac{\angle B}{2}=\frac{\angle C}{2}$.

Since $BO$ and $CO$ are bisectors of $\angle B$ and $\angle C$, we also have

$\angle ABO=\frac{\angle B}{2}$ and $\angle ACO=\frac{\angle C}{2}$.

$\angle ABO=\frac{\angle B}{2}=\frac{\angle C}{2}=\angle ACO$.

Consider $△BCO$:

$\angle OBC=\angle OCB$

$\Rightarrow BO=CO$ ....... [Sides opposite to equal angles are equal]

Finally, consider triangles $ABO$ and $ACO$.

$BA=CA$ ...... (given);

$BO=CO$ ...... (proved);

$\angle ABO=\angle ACO$ (proved).

Hence, by $\text{S.A.S}$ postulate

$△ABO\cong △ACO$

$\Rightarrow \angle BAO=\angle CAO\Rightarrow AO$ bisects $\angle A$.