In the given figure, I is the incenter of △ABC, BI when produced meets the circumcircle of △ABC at D. Given, ∠BAC = 55° and ∠ACB = 65°; calculate :

(i) ∠DCA,

(ii) ∠DAC,

(iii) ∠DCI,

(iv) ∠AIC.

(i) Join IA, IC and CD.

In △ABC,

⇒ ∠ABC + ∠BAC + ∠ACB = 180°

⇒ ∠ABC + 55° + 65° = 180°

⇒ ∠ABC + 120° = 180°

⇒ ∠ABC = 180° - 120° = 60°.

IB is the bisector of ∠ABC [As I is the incenter].

∠ABD = $\dfrac{1}{2}$∠ABC = $\dfrac{1}{2}$ x 60° = 30°.

We know that,

Angle in same segment are equal.

∴ ∠DCA = ∠ABD = 30°.

Hence, ∠DCA = 30°.

(ii) ∠CBD = ∠ABD = 30° [As IB is bisector of ∠ABC]

We know that,

Angle in same segment are equal.

∴ ∠DAC = ∠CBD = 30°.

Hence, ∠DAC = 30°.

(iii) As CI is the angle bisector of ∠ACB.

∠ACI = $\dfrac{1}{2}$∠ACB = $\dfrac{1}{2} \times 65°$ = 32.5°

From figure,

∠DCI = ∠DCA + ∠ACI = 30° + 32.5° = 62.5°

Hence, ∠DCI = 62.5°.

(iv) As AI is the angle bisector of ∠BAC.

∠IAC = $\dfrac{1}{2}$∠BAC = $\dfrac{1}{2} \times 55°$ = 27.5°

In △AIC,

⇒ ∠IAC + ∠ACI + ∠AIC = 180°

⇒ 27.5° + 32.5° + ∠AIC = 180°

⇒ 60° + ∠AIC = 180°

⇒ ∠AIC = 180° - 60° = 120°.

Hence, ∠AIC = 120°.