In the figure (ii) given below, I is the incentre of △ABC. AI produced meets the circumcircle of △ABC at D. Given that ∠ABC = 55° and ∠ACB = 65°, calculate

(i) ∠BCD

(ii) ∠CBD

(iii) ∠DCI

(iv) ∠BIC.

(i) Join BI and CI as shown in the figure below:

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 = $\dfrac{60°}{2}$ = 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 = $\dfrac{65°}{2} = 32\dfrac{1}{2}°$.

From figure,

∠DCI = ∠BCD + ∠BCI = $30° + 32\dfrac{1}{2}° = 62\dfrac{1}{2}°$.

Hence, the value of ∠DCI = $62\dfrac{1}{2}°$.

(iv) ∠IBC = $\dfrac{55°}{2} = 27\dfrac{1}{2}°$

∠ICB = $\dfrac{65°}{2} = 32\dfrac{1}{2}°$

∠BIC = 180° - (∠IBC + ∠ICB)

$= 180° - \Big(\dfrac{55°}{2} + \dfrac{65°}{2}\Big) \\[1em] = 180° - \Big(\dfrac{120°}{2}\Big) \\[1em] = 180° - 60° \\[1em] = 120°.$

Hence, the value of ∠BIC = 120°.