Question

In the given figure, BI is the bisector of ∠ABC and Cl is the bisector of ∠ACB. Find ∠BIC.

Answer 1

In Δ ABC,

BI is the bisector of ∠ABC and CI is the bisector of ∠ACB.

∵ AB = AC

∴ ∠B = ∠C ............(Angles opposite to equal sides)

But ∠A = 40°
and ∠A + ∠B + ∠C = 180° ......(Angles of a triangle)

⇒ 40° + ∠B + ∠B = 180°

⇒ 40° + 2 ∠B = 180°

⇒ 2 ∠B = 180° − 40° = 140°

⇒ ∠B =`(140°)/2=70°`

∴ ∠ABC = ∠ACB = 70°

But BI and CI are the bisectors of ∠ABC and ∠ACB respectively.

∴ ∠IBC =`1/2` ∠ABC =`1/2(70°)=35°`

and ∠ICB =`1/2` ∠ACB =`1/2xx70°=35°`

Now in Δ IBC,

∠BIC + ∠IBC + ∠ICB = 180° ........(Angles of a triangle)

⇒ ∠BIC + 35° + 35° = 180°

⇒ ∠BIC + 70° = 180°

⇒ ∠BIC = 180° − 70° = 110°

Hence ∠BIC = 110°