#### Question

If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC = 80°. Calculate :

(i) ∠DBC (ii) ∠IBC (iii) BIC

#### Solution

Join DB and DC, IB and IC

∠BAC 66°, ∠ABC 80°, I is the incentre of the ΔABC,

(i) since ∠DBC and ∠DACare in the same segment

∠DBC = ∠DAC

But, `∠DAC = 1 /2 ∠BAC = 1 /2 xx 66° = 33°` =

∴ ∠DBC = 33°

(ii) Since I is the incentre of ∆ABC, IB bisects ∠ABC

`∴ ∠IBC = 1/2 ∠ABC = 1 /2 xx 80° = 40° `

(iii) ∠BAC = 66° and ∠ABC = 80°

In ∠ABC, ∠ACB = 180° - (∠ABC + ∠BAC)

⇒ ∠ACB = 180° - (80° + 66°)

⇒ ∠ACB = 180° - (156°)

⇒ ∠ACB = 34°

Since IC bisects the ∠C

`∴∠ICB = 1/2 ∠C = 1/2xx 34° = 17° `

Now in IBC

∠IBC + ∠ICB + ∠BIC =180°

⇒ 40° +17° + ∠BIC = 180°

⇒ 57° + ∠BIC = 180°

⇒ ∠BIC = 180° - 57°

⇒ ∠BIC = 123°