In ∆ABC, AB = AC and the bisectors of ∠B and ∠C meet at a point O.


In ABCAB = 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.


In triangle ABC, we have:
AB = AC   (Given)

$\Rightarrow \angle B=\angle C$

$\Rightarrow \frac{1}{2} \angle B=\frac{1}{2} \angle C$

$\Rightarrow \angle O B C=\angle O C B$

$\Rightarrow B O=C O$

Now, in $\triangle A O B$ and $\triangle A O C$, we have:

$O B=O C \quad$ (Proved)

$A B=A C \quad$ (Given)

$A O=A O \quad$ (Common)

$\therefore \triangle A O B \cong \triangle A O C \quad$ (SSS criterion)

i.e., $\angle B A O=\angle C A O$ (Corresponding angles of congruent triangles)

So, it shows that ray AO is the bisector of A∠A.
Hence, proved.


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