$ABC$ is an isosceles triangle with $AB=AC$. Bisectors of $\angle B$ and $\angle C$ meet at $O$. Join $AO$. Prove that,$OB=OC$ and $AO$ bisect $\angle A$.
Given :
$ABC$ is an isosceles triangle.
$AB=AC$
Bisectors of angles $B$ and $C$ intersect at $O$.
To do :
We have to prove that $OB=OC$ and $AO$ bisect $\angle A$.
Solution :
$OB$ is the bisector of $∠B$.
This implies,
$∠ABO = ∠OBC = \frac{1}{2} ∠B$
$OC$ is the bisector of $∠C$.
This implies,
$∠ACO = ∠OCB = \frac{1}{2} ∠C$
We know that,
Angles opposite to equal sides are equal.
Sides opposite to equal angles are equal.
Therefore,
$∠ACB = ∠ABC$
$ \frac{1}{2} ∠ACB = \frac{1}{2} ∠ABC$
$∠OCB = ∠OBC$
$OB=OC$
In triangles $AOB$ and $AOC$,
$OB=OC$
$AB=AC$ (Given)
$AO=AO$ (Common side)
Therefore,
Triangle $AOB$ is congruent to triangle $AOC$.
This implies,
$∠OAB = ∠OAC$ (CPCT)
Hence proved.
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