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, $\angle\text{MOC}-=\angle\text{ABC}$
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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