In the given figure, $\angle BAC=30^o$. Show that $BC$ is equal to the radius of the circumcircle of $△ABC$ whose centre is $O$.

Join the lines $AB$ and $OC$

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference

So we get

It is given that $\angle BAC=30^o$

By multiplication 

In $△BOC$

We know that the radii are equal 

The base angles of an isosceles triangle are equal

Consider $△BOC$

Using the angle sum property

By substituting the value we get

So we get

By Subtraction

By division 

We get $\angle OBC=\angle OCB=\angle BOC=60^o$

We know that $△BOC$ is an equilateral triangle

So we get $OB=OC=BC$

Therefore, it is proved that $BC$ is equal to the radius of the circumcircle of $△ABC$ whose centre is $O$.