Question

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

Answer 1

Proved below.

Step-by-step explanation:

Given:

In a Δ ABC, ∠ABC =∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O. Also ∠BOC = 120°.

To prove:

∠A =∠B =∠C = 60°.

Proof:

Let ABC be a triangle and BO and CO be the bisectors of the base angle of ∠ABC and ∠ACB respectively.

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

$\angle BOC = 90 +\frac{1}{2}\angle A$

$120 = 90 +\frac{1}{2}\angle A$

$120 -90 =\frac{1}{2}\angle A$

$30 =\frac{1}{2}\angle A$

$\angle A= 30 \times 2$

∠A = 60°                                [1]

Also ∠B and ∠C are equal as it is given that ∠ABC =∠ACB.

∠A+∠B+∠C=180°     [Sum of three angles of a triangle is 180°]

⇒60°+2∠B=180°         [∵ ∠ABC=∠ACB]

⇒∠B = 60° = ∠C                   [2]

From Eq (1) and (2), we get

Hence ∠A =∠B =∠C = 60°

Hence proved.