O is the circumecentre of `Delta ABC and D` is the mid point of BC. If `angle BAC=40^(@)`, then find the value of `angle BOD`.

Explanation

D is the mid-point of `BC, :. BD=CD`
Now, in triangles `Delta BOD and Delta COD, OB= OC` [radii of the same circle]
`BD=CD and angleOBD= angleOCD [ :. =OC]`
`:. Delta BOD ~=Delta COD` [ by the conditon of S-A-S congruent]
`:. angleBOD= angleCOD [ :.` smiliar angles of congruent triangles]
`:. angleBOD=(1)/(2) angle.........(1) [ :. angleBOC=angle BOD + angleCOD= angleBOD+ angleBOD= 2 angleBOD]`
Now,the central angle produced by the are ` overset (frown) (BC)= angleBOC` and its angle is circle `=angleBAC`.
`:. angleBOC= 2 angleBAC`[ by theorem-34]
`=2xx40^(@) [ :. angleBAC=40^(@)]=80^(@)`
The from (1) we get, `angleBOD=(1)/(2) angleBOC=(1)/(2)xx80^(@) [ :. angleBOC=80^(@)]=40^(@)`