In $Δ A B C, A B = A C$ , and the bisectors of $∠ B$ and $∠ C$ intersect at point $O$ . Prove that $B O = O C$ and the ray $A O$ is the bisector of $∠ B A C$ .

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Solution

Given $O C, O B$ bisect $∠ B, ∠ C$
$A B = A C$
To prove : $A D$ bisect $∠ A$
Proof : $∠ B = ∠ C$
as $A B = A C$
$\frac{1}{2} ∠ B = \frac{1}{2} ∠ C$
i.e., $∠ 2 = ∠ 4, ∠ 1 = ∠ 3$
$O B = O C$
In $Δ A O C, Δ A O B$
$A C = A B$
$∠ 1 = ∠ 3$
$O B = O C$
$∴ Δ A O C ≅ Δ A O D$ (By SAS)
$∴$ AO bisect $∠ A$

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