Question

In a $△ABC$, it is given that $AB=BC$ and the bisectors of $\angle B$ and $\angle C$ intersect at O. If M is a point on BO produced, prove that $\angle MOC=\angle ABC$

Answer 1

REF.Image

As angles opposite to two equal

sides of a triangle are equal.

$\Rightarrow \angle ABC=\angle ACB(AB=AC)$

$\Rightarrow \angle B=\angle C$

Given, OB and OC are bisector of $\angle B,\angle C$

In $△OBC$,

$\angle BOC+\frac{\angle B}{2}+\frac{\angle C}{2}={180}^{\circ }$

$\Rightarrow \angle BOC+2\times \left(\frac{\angle B}{2}\right)={180}^{\circ }$ $(\angle B=\angle C)$

$\Rightarrow \angle BOC+\angle B={180}^{\circ }$

$\Rightarrow \angle BOC={180}^{\circ }-\angle B$

As $\angle MOC+\angle BOC={180}^{\circ }$ (Linear angles)

$\Rightarrow \angle MOC={180}^{\circ }-\angle BOC$

$\Rightarrow \angle MOC=180-(180-\angle B)$

$\Rightarrow \angle MOC=180-180+\angle B$

$\Rightarrow \angle MOC=\angle B$

$\Rightarrow \angle MOC=\angle ABC$