O is the circumcentre of `Delta ABC` and OD is perpendicular on the side BC, prove that ` angle BOD= angle BAC`

Explanation

The circumcentre of `Delta ABC` is O and `OD bot BC`.
To prove : We have to prove that `angle BOD= angle BAC`
Construction: let us join `O,A, O,B and O,C`
Proof : In `Delta ABOD and Delta COD, OB= Oc [ :. ` radii of same circle,]
`angle BOD= angle OCD[ :. OB= OC] and`
`angle BDO= angle CDO[ :.` each is right angle]
`:. Delta BOD~= Delta COD` [ by the condition of A-A-S congruence]
`:. angle BOD= angle COD[ :. ` similar angles of congruent triangles]
`:. angleBOC= angle BOD+ angle COD= angle BOD+ angle BOD [ :. angle COD= angle BOD]`
`2 angle BOD ......(1)`

Now, the central angle produed by the chrod `BC= angle BOC` and angle in circle `= angle BAC.`
`:. angle BOC=2 angle BAC` [ by theorem-34] ......(2)
Then adding (1) and (2) we get,
`2 angle BOD=2 angle BAC or, angle BOD= angle BAC`
Hence `angle BOD= angle BAC`