Question

In ∆ABC, AB = AC and the bisectors of ∠B and ∠C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠A.

Answer 1

In triangle ABC, we have:
AB = AC (Given)
$$\begin{gathered} \Rightarrow \angle B=\angle C \\ \Rightarrow \frac{1}{2}\angle B=\frac{1}{2}\angle C \\ \Rightarrow \angle OBC=\angle OCB \\ \Rightarrow BO=CO \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{in}△AOB\mathrm{and}△AOC,\mathrm{we}\mathrm{have}: \\ OB=OC\left(\mathrm{Proved}\right) \\ AB=AC\left(\mathrm{Given}\right) \\ AO=AO\left(\mathrm{Common}\right) \\ \therefore △AOB\cong △AOC(\mathrm{SSS}\mathrm{criterion}) \end{gathered}$$
i.e., $\angle BAO=\angle CAO(\mathrm{Corresponding}\mathrm{angles}\mathrm{of}\mathrm{congruent}\mathrm{triangles})$

So, it shows that ray AO is the bisector of $\angle A$.
Hence, proved.