# In an isosceles triangle $ABC$, with $AB = AC$, the bisectors of $\angle B = \angle C$ intersect each other at $O$.Join $A$ to $O$. Show that $(i) OB = OC $ and $(ii) AO$ bisects $\angle A$.

Answer

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Hint: Join $A$ to $O$. Apply the given condition to the isosceles triangle and use similarity criterion.

According to given data we have 3 conditions,

$AB = AC \to (1)$

$OB$ is the bisector of $\angle B$

So,$\angle ABO = \angle OBC = \dfrac{1}{2}\angle B \to (2)$

$OC$ is the bisector of$\angle C$

So,$\angle ACO = \angle OCB = \dfrac{1}{2}\angle C$$ \to (3)$

Case-1

So, here we have to prove $OB = OC$

Proof:

Now by using condition (1) we can say that,

$AB = AC$

From this condition we say that

$ \Rightarrow \angle ACB = \angle ABC$ [ Where we know that Angles opposite to equal sides are equal]

$\

\Rightarrow \dfrac{1}{2}\angle ACB = \dfrac{1}{2}\angle ABC \\

\\

\ $

$ \Rightarrow \angle OCB = \angle OBC$ [From (2) and (3)]

Hence,

$OB = OC$ [Sides opposite to equal angles are equal]

Hence proved that $OB = OC$.

Case-2

We have to prove that $\angle OAB = \angle OAC$

By using case (1) we know that $OB = OC$

And also from $\Delta ABO$ and$\Delta ACO$, we have

$ \Rightarrow AB = AC$ (Given)

$ \Rightarrow AO = OA$ (Common)

$ \Rightarrow OB = OC$(From (case 1))

$\therefore \Delta ABO \cong \Delta ACO$ (By SSS Congruence rule)

$ \Rightarrow \angle OAB = \angle OAC$ (CPCPT Theorem)

Hence we have proved that$\angle OAB = \angle OAC$.

CASE - 1 CASE - 2

NOTE: In this problem given construction is mandatory to prove the given statements so join $A$ and $O$ points.

According to given data we have 3 conditions,

$AB = AC \to (1)$

$OB$ is the bisector of $\angle B$

So,$\angle ABO = \angle OBC = \dfrac{1}{2}\angle B \to (2)$

$OC$ is the bisector of$\angle C$

So,$\angle ACO = \angle OCB = \dfrac{1}{2}\angle C$$ \to (3)$

Case-1

So, here we have to prove $OB = OC$

Proof:

Now by using condition (1) we can say that,

$AB = AC$

From this condition we say that

$ \Rightarrow \angle ACB = \angle ABC$ [ Where we know that Angles opposite to equal sides are equal]

$\

\Rightarrow \dfrac{1}{2}\angle ACB = \dfrac{1}{2}\angle ABC \\

\\

\ $

$ \Rightarrow \angle OCB = \angle OBC$ [From (2) and (3)]

Hence,

$OB = OC$ [Sides opposite to equal angles are equal]

Hence proved that $OB = OC$.

Case-2

We have to prove that $\angle OAB = \angle OAC$

By using case (1) we know that $OB = OC$

And also from $\Delta ABO$ and$\Delta ACO$, we have

$ \Rightarrow AB = AC$ (Given)

$ \Rightarrow AO = OA$ (Common)

$ \Rightarrow OB = OC$(From (case 1))

$\therefore \Delta ABO \cong \Delta ACO$ (By SSS Congruence rule)

$ \Rightarrow \angle OAB = \angle OAC$ (CPCPT Theorem)

Hence we have proved that$\angle OAB = \angle OAC$.

CASE - 1 CASE - 2

NOTE: In this problem given construction is mandatory to prove the given statements so join $A$ and $O$ points.

Last updated date: 22nd Sep 2023

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