In the given figure, I is the incentre of `Delta ABC` and AI when produced meets the circumcentre of `Delta ABC` in D. If `/_ BAC = 56^(@) ` and `/_ACB = 64^(@)` ,calculate
(i) `/_ DBC , (ii) /_ IBC, (iii) /_ ADB ` and `(iv) /_ BID `

Explanation

Join BI.
In `Delta BAC`,we have
`/_BAC + /_ ACB + /_ ABC = 180^(@)` [sum of the `/_ s` ofa `Delta `]
`implies 56^(@) + 64^(@) + /_ ABC = 180^(@)`
`implies /_ ABC = 180^(@) - 120^(@) = 60^(@)`
Now, `/_ DAC = (1)/(2) /_ BAC = ((1)/(2) xx 56^(@)) = 28^(@)` `[ :'` AD is the bisector of `/_ A ` ].
(i) `/_ DBC = /_ DAC = 28^(@)`
(ii) `/_ IBC = (1)/(2) /_ ABC = ((1)/(2) xx 60^(@))= 30^(@) ` [`:' ` BI is the bisector of `/_ ABC`] .
(iii) `/_ ADB = /_ ACB = 64^(@)` [angles in the segment ].
(iv) In `Delta BDI`, we have
`/_ DBI + /_ BDI + /_ BID = 180^(@)` [ sum of the `/_ s` of a triangle ]
`implies /_ DBC + /_ IBC + /_ ADB + /_ BID =180^(@)`
`[ :' /_ DBI =/_ DBC + /_ IBC` and `/_ BDI = /_ ADB ]`
`implies 28^(@) + 30^(@) + 64^(@) + /_ BID = 180^(@)`
`implies /_ BID = 180^(@) -122^(@)=58^(@)`