Question

The bisectors $$\angle B$$ and $$\angle C$$ of and isosceles triangle with $$AB = AC$$ intersect each other at a point $$O$$. $$BO$$ is produced to meet $$AC$$ at a point $$M$$. Prove that $$\angle MOC = \angle ABC$$.

Answer 1

Consider $$\triangle ABC$$

It is given that $$AB = AC$$

So we get

$$\angle ABC = \angle ACB$$

Dividing both sides by $$2$$ we get

$$\dfrac {1}{2} \angle ABC = \dfrac {1}{2} \angle ACB$$

So we get

$$\angle OBC = \angle OCB$$

By using the exterior angle property

We get

$$\angle MOC = \angle OBC + \angle OCB$$

We know that $$\angle OBC = \angle OCB$$

So we get

$$\angle MOC = 2 \angle OBC$$

We know that $$OB$$ is the bisector of $$\angle ABC$$

$$\angle MOC = \angle ABC$$

Therefore, it is proved that $$\angle MOC = \angle ABC$$.