In the given figure, I is the incentre of Δ ABC. Here AI produced meets the circumcircle of Δ ABC at D. If ∠ABC = 55° and ∠ACB = 65°, calculate :

(i) ∠BCD

(ii) ∠CBD

(iii) ∠DCI

(iv) ∠BIC

(i) Join BI and CI.

In △ABC,

⇒ ∠BAC + ∠ABC + ∠ACB = 180° (∵ sum of angles = 180°.)

⇒ ∠BAC + 55° + 65° = 180°

⇒ ∠BAC + 120° = 180°

⇒ ∠BAC = 180° - 120°

⇒ ∠BAC = 60°.

I is the incentre,

∴ I lies on the bisectors of angle of the △ABC,

∴ ∠BAD = ∠CAD = $\dfrac{60^{\circ}}{2}$ = 30°.

∠BCD = ∠BAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠BCD = 30°.

(ii) Similarly,

∠CBD = ∠CAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠CBD = 30°.

(iii) The line CI bisects ∠C (∵ I lies on the bisectors of angle of the △ABC).

∴ ∠BCI = $\dfrac{65^{\circ}}{2} = 32 \dfrac{1}{2}^{\circ}$.

From figure,

∠DCI = ∠BCD + ∠BCI = 30° + $32 \dfrac{1}{2}^{\circ}$

= 62.5°.

Hence, the value of ∠DCI = 62.5°.

(iv) ∠IBC = $\dfrac{55}{2}^{\circ}$ = 27.5°

∠ICB = $\dfrac{65}{2}^{\circ}$ = 32.5°

In triangle BIC,

By angle sum property of triangle,

∠BIC + ∠IBC + ∠ICB = 180°

∠BIC = 180° - (∠IBC + ∠ICB)

= 180° - (27.5° + 32.5°)

= 180° - 60°

= 120°.

Hence, the value of ∠BIC = 120°.