Question

Bisectors of angles $B$ and $C$ of an isosceles triangle intersect each other at $O$ and $AB=AC$. $BO$ is produced to a point $M$ on side $AC$. Prove that $\angle MOC=\angle ABC$.

Answer 1

Given:

Lines $OB$ and $OC$ are the bisectors of $\angle B$ and $\angle C$ of an isosceles $\Delta ABC$ such that $AB=AC$ which intersect each other at $O$ and $BO$ is produced to $M$.

To prove:

$\angle MOC=\angle ABC$

Consider the diagram shown below.

Proof:

In $\Delta ABC$,

$AB=AC$ (given)

$\angle ACB=\angle ABC$ (angles opposite to equal sides are equal)

$\frac{1}{2}\angle ACB=\frac{1}{2}\angle ABC$ (dividing both sides by 2)

Therefore,

$\angle OCB=\angle OBC$ …… (1)

(Since, $OB$ and $OC$ are the bisector of $\angle B$ and $\angle C$)

Now, from equation (1), we have

$\angle MOC=\angle OBC+\angle OCB$ $\because$ sum of interior angle = Exterior angle

$\angle MOC=2\angle OBC$

$\Rightarrow \angle MOC=2\angle ABC\times \frac{1}{2}=\angle ABC$

(Since, $OB$ is the bisector of $\angle B$)

Hence, proved.