Question

In a ΔABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°. 

Answer 1

Given,
In ΔABC 

∠ABC=∠ ACB 

Divide both sides by '2' 

`1/2∠ABC=1/2∠ACB` 

⇒ ∠OBC=∠ OCB              [∵ OB, OC bisects ∠B and ∠C] 

 

Now 

`∠BOC=90^@+1/2∠A` 

`120^@-90^@=1/2∠A` 

`30^@xx(2)=∠A` 

`∠A=60^@` 

Now in Δ ABC 

`∠A+∠ABC+∠ACB=180^@`      (Sum of all angles of a triangle) 

                                                            [∵∠ABC=∠ACB]

`60^@+2∠ABC=180^@` 

`2∠ABC=180^@-60^2` 

`∠ABC=120^@/2=90^@` 

`∠ABC=∠ ACB` 

`∴ ∠ACB=60^@` 

Hence proved.