Question

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:

1. ∠DBC,
2. ∠IBC,
3. ∠BIC.

Answer 1

Join DB and DC, IB and IC

∠BAC = 66°, ∠ABC = 80°,

I is the incentre of the ΔABC,

i. Since ∠DBC and ∠DAC are in the same segment,

∠DBC = ∠DAC

But, `∠DAC = 1/2 ∠BAC`

= `1/2 xx 66^circ`

= 33°

∴ ∠DBC = 33°

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

∴ `∠IBC = 1/2 ∠ABC`

= `1/2 xx 80^circ`

= 40°

iii. ∴ ∠BAC = 66° and ∠ABC = 80°

In ΔABC, ∠ACB = 180° – (∠ABC + ∠BAC)

`=>` ∠ACB = 180° – (80° + 66°)

`=>` ∠ACB = 180° – (156°)

`=>` ∠ACB = 34°

Since IC bisects the ∠C

∴ `∠ICB = 1/2 ∠C`

= `1/2 xx 34^circ`

= 17°

Now in ΔIBC

∠IBC + ∠ICB + ∠BIC =180°

`=>` 40° + 17° + ∠BIC = 180°

`=>` 57° + ∠BIC = 180°

`=>` ∠BIC = 180° – 57°

`=>` ∠BIC = 123°