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.

Join DB and DC, IB and IC.

(i) From figure,

∠DAC = $\dfrac{1}{2}$∠BAC [As I is the incenter]

∠DAC = $\dfrac{1}{2}$ x 66° = 33°.

As angle in same segment are equal.

∴ ∠DBC = ∠DAC = 33°.

Hence, ∠DBC = 33°.

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

∴ ∠IBC = $\dfrac{1}{2}$∠ABC

= $\dfrac{1}{2} \times 80°$ = 40°.

Hence, ∠IBC = 40°.

(iii) In ∆ABC,

⇒ ∠ACB + ∠ABC + ∠BAC = 180° [By angle sum property]

⇒ ∠ACB = 180° - ∠ABC - ∠BAC

⇒ ∠ACB = 180° - 80° - 66°

⇒ ∠ACB = 180° - 146°

⇒ ∠ACB = 34°.

As I is incenter so, IC bisects ∠C

∴ ∠ICB = $\dfrac{1}{2}$∠ACB = $\dfrac{1}{2} \times 34°$ = 17°.

In ∆IBC

⇒ ∠IBC + ∠ICB + ∠BIC = 180° [By angle sum property of triangle]

⇒ 40° + 17° + ∠BIC = 180°

⇒ 57° + ∠BIC = 180°

⇒ ∠BIC = 180° - 57° = 123°.

Hence, ∠BIC = 123°.