Question
In a $\triangle\text{ABC},$ it is given that $\text{AB}=\text{AC}$ and the bisectors of $B$ and $C$ intersect at $O$. If $M$ is a point on $BO$ produced, prove that $\angle\text{MOC}=\angle\text{ABC}.$
✓
Answer
Given that in $\triangle\text{ABC},$
$\text{AB}=\text{AC}$ and the bisector of $\angle\text{B}$ and $\angle\text{C}$ intersect at $O$.
If $M$ is a point on $BO$ produced
We have to prove $\angle\text{MOC}=\angle\text{ABC}$
Since, $\text{AB}=\text{AC}$ ABC is isosceles $\angle\text{B}=\angle\text{C}\text{ (or})$
$\angle\text{ABC}=\angle\text{ACB}$
Now, $BO$ and $CO$ are bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ respectively $\Rightarrow\text{ABO}=\angle\text{OBC}=\angle\text{ACO}=\angle\text{OCB}$
$=\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ACB}\ ...(\text{i)}$
We have, $\triangle\text{OBC}$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ\ ....(\text{ii)}$ And also $\angle\text{BOC}+\angle\text{COM}=180^\circ\ ....(\text{iii)}$ [Straight angle]
Equating $(ii)$ and $(iii)$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=\angle\text{BOC}+\angle\text{MOC}$
$\angle\text{OBC}+\angle\text{OCB}=\angle\text{MOC}$ [From $(i)]$
$2\Big(\frac{1}{2}\angle\text{ABC}\Big)=\angle\text{MOC}$ [From $(i)]$
$\angle\text{ABC}=\angle\text{MOC}$
Therefore, $\angle\text{MOC}-=\angle\text{ABC}$
$\text{AB}=\text{AC}$ and the bisector of $\angle\text{B}$ and $\angle\text{C}$ intersect at $O$.
If $M$ is a point on $BO$ produced
We have to prove $\angle\text{MOC}=\angle\text{ABC}$
Since, $\text{AB}=\text{AC}$ ABC is isosceles $\angle\text{B}=\angle\text{C}\text{ (or})$
$\angle\text{ABC}=\angle\text{ACB}$
Now, $BO$ and $CO$ are bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ respectively $\Rightarrow\text{ABO}=\angle\text{OBC}=\angle\text{ACO}=\angle\text{OCB}$
$=\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ACB}\ ...(\text{i)}$
We have, $\triangle\text{OBC}$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ\ ....(\text{ii)}$ And also $\angle\text{BOC}+\angle\text{COM}=180^\circ\ ....(\text{iii)}$ [Straight angle]
Equating $(ii)$ and $(iii)$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=\angle\text{BOC}+\angle\text{MOC}$
$\angle\text{OBC}+\angle\text{OCB}=\angle\text{MOC}$ [From $(i)]$
$2\Big(\frac{1}{2}\angle\text{ABC}\Big)=\angle\text{MOC}$ [From $(i)]$
$\angle\text{ABC}=\angle\text{MOC}$
Therefore, $\angle\text{MOC}-=\angle\text{ABC}$
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