In , AB = AC and the bisectors of and meet at a point O. Prove that BO = CO and the ray AO is the bisector of .

Given: A in which AB = AC, BO and CO are bisectors of and To Prove: In , we have, =1 2 and, =1 2 But, = [ AB = AC (given)] SO, = Since base angle are equal, sides are equal = OC...(1) Since OB and OC are the bisectors of angles, and respectively, we have =1 2 =1 2 = ...(2) Now, in and AB = AC [Given] = [from (2)] BO = OC [from (1)] Thus, by Side-Angle-side criterion of congruence, we have [BY SAS] The corresponding parts of the congruent triangles are equal = [By C.P.C.T.] i.e. AO bisects .

In , AB = AC and the bisectors of and meet at a point O. Prove th…

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Question

In $\triangle\text{ABC,}$ $AB = AC$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at a point $O$.
Prove that $BO = CO$ and the ray $AO$ is the bisector of $\angle\text{A}.$

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Answer

Given: A $\triangle\text{ABC}$ in which $AB = AC, BO$ and $CO$ are bisectors of $\angle\text{B}$ and $\angle\text{C}$

To Prove: In $\triangle\text{BOC},$
we have, $\angle\text{OBC}=\frac{1}{2}\angle\text{B}$ and,
$\angle\text{OBC}=\frac{1}{2}\angle\text{C}$ But, $\angle\text{B}=\angle\text{C}$ [$\therefore$ $AB = AC$ (given)]
SO, $\angle\text{OBC}=\angle\text{OCB}$
Since base angle are equal, sides are equal
$\Rightarrow\text{OB = OC}...(1)$
Since $OB$ and $OC$ are the bisectors of angles,
$\angle\text{B}$ and $\angle\text{C}$ respectively,
we have $\angle\text{ABO}=\frac{1}{2}\angle\text{B}$
$\angle\text{ACO}=\frac{1}{2}\angle\text{C}$
$\Rightarrow\angle\text{ABO}=\angle\text{ACO}...(2)$
Now, in $\triangle\text{ABO}$ and $\triangle\text{ACO}$
$\text{AB = AC}$ [Given] $\angle\text{ABO}=\angle\text{ACO}$ [from $(2)$]
$\text{BO = OC}$ [from $(1)$] Thus, by Side-Angle-side criterion of congruence,
we have $\triangle\text{ABO}\cong\triangle\text{ACO}$ [BY $SAS$]
The corresponding parts of the congruent triangles are equal
$\therefore\angle\text{BAO}=\angle\text{CAO}$ [By $C.P.C.T.$] i.e. $AO$ bisects $\angle\text{A}.$