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

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Solution

In $△$ ABC, we have
AB = AC

$\Rightarrow ∠ B = ∠ C$ [Angles opposite to equal sides are equal]
$\Rightarrow \frac{1}{2} ∠ B = \frac{1}{2} ∠ C$
$\Rightarrow ∠ O B C = ∠ O C B; ∠ O B A = ∠ O C A . . (1)$
[OB and OC are bisector of $∠ B a n d ∠ C$ respectively]
$\Rightarrow O B = O C . . (2)$
[Sides opp. to equal $∠ s$ are equal]
Now, in $△$ ABO and $△$ ACO, we have
AB = AC [Given]
OA = OA [Common]
OB = OC [From (2)]
So, by SSS criterion of congruence
$△ A B O ≅ △ A C O \Rightarrow ∠ B A O = ∠ C A O$
[Corresponding parts of congruent triangles are equal]
$\Rightarrow$ AO is the bisector of $∠$ BAC.

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In △ABC, AB=AC and the bisectors of angles B and C intersect at point O. Prove that BO = CO and the ray AO is the bisector of angle BAC.

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