Question

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

Answer 1

Let ABC be a triangle and BO and CO be the bisectors of the base anglerespectively.

We know that if the bisectors of angles ∠ABC and ∠ACB of a triangle ABC meet at a point O, then

$\angle BOC=90°+\frac{1}{2}\angle A$

$$\begin{gathered} \therefore 120°=90°+\frac{1}{2}\angle A \\ \Rightarrow 30°=\frac{1}{2}\angle A \\ \Rightarrow \angle A=60° \end{gathered}$$

are equal as it is given that .

$$\begin{gathered} \angle A+\angle B+\angle C=180°\left[\mathrm{Sum}\mathrm{of}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°\right] \\ \Rightarrow 60°+2\angle B=180°\left[\because \angle ABC=\angle ACB\right] \\ \Rightarrow \angle B=60° \end{gathered}$$

Hence, .