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°.

Question

In a &#x394; ABC , &#x2220; ABC = &#x2220; ACB and the bisectors of &#x2220; ABC and &#x2220; ACB intersect at O such that &#x2220; BOC = 120&#xB0;. Shoe that &#x2220; A = &#x2220; B = &#x2220; C = 60&#xB0;.

Philoid · Accepted Answer

Given,

∠ ABC = ∠ ACB

Divide both sides by 2, we get

∠ABC = ∠ACB

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

Now,

∠BOC = 90^o + ∠A

120^o – 90^o = ∠A

30^o * 2 = ∠A

∠A = 60^o

Now in

∠A + ∠ABC + ∠ACB = 180^o [Sum of all angles of a triangle]

60^o + 2∠ABC = 180^o [Therefore, ∠ABC = ∠ACB]

2∠ABC = 180^o – 60^o

2∠ABC = 120^o

∠ABC = 60^o

Therefore, ∠ABC = ∠ACB = 60^o

Hence, proved

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°.