If '$$O$$' is the circumcentre of a triangle $$ABC$$ and $$OD$$ is perpendicular to $$BC$$, then find $$\angle BOD$$.

In triangle $$BOC$$ and triangle $$COD$$

$$BO$$ = $$CO$$   ------ (Both are radius)

$$\angle BDO$$ = $$\angle CDO$$ = $$90^\circ$$

$$OD$$ is common in both triangle 

So,$$\angle BOD$$ = $$\angle COD$$ --------- i

As we know that angle at centre is double the angle at the circumference

$$\angle BOC$$ = 2$$\angle BAC$$

2$$\angle BOD$$ = 2$$\angle BAC$$  ------ ( from i)

$$\angle BOD$$ = $$\angle BAC$$ 

Hence, $$\angle BOD$$ = $$\angle A$$