Question

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

1. ∠DCA,
2. ∠DAC,
3. ∠DCI,
4. ∠AIC.

Answer 1

Join AD, DC, AI and CI,

In ΔABC,

∠BAC = 55°, ∠ACB = 65°

∴ ∠ABC = 180° – (∠BAC + ∠ACB)

= 180° – (55° + 65°)

= 180° – 120°

= 60° 

In cyclic quad. ABCD,

∠ABC + ∠ADC = 180° 

`\implies` 60° + ∠ADC = 180° 

∴ ∠ADC = 180° – 60° = 120°

In ΔADC,

∠DAC + ∠DCA + ∠ADC = 180°

`\implies` ∠DAC + ∠DCA + 120° = 180°

`\implies` ∠DAC + ∠DCA = 180° – 120° = 60°

But ∠DAC = ∠DCA

(I lies on the bisector of ∠ABC)

∴ ∠DAC = ∠DCA = 30°

∴ DI is perpendicular bisector of AC

∠AIC = ∠ADC = 120°

∴ IC is the bisector of ∠ACB

∴ ∠ICA = `65^circ/2` = 32.5°

∴ ∠DCI = ∠DCA + ∠ACI

= 30° + 32.5°

= 62.5°.

= (62.5)°

= 60° 30'.