In a Δ ABC, ∠ABC =∠ACB and the bisectors of ∠ABC and ∠ACB int

1.	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» Given, In ΔABC ∠ ABC = ∠ ACB Divide both sides by 2, we get \(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\)∠ACB ∠OBC = ∠OCB [Therefore, OB, OC bisects ∠B and ∠C] Now, ∠BOC = 90° + \(\frac{1}{2}\)∠A 120° – 90° = \(\frac{1}{2}\)∠A 30° x 2 = ∠A ∠A = 60° Now in ΔABC ∠A + ∠ABC + ∠ACB = 180°[Sum of all angles of a triangle] 60° + 2∠ABC = 180°[Therefore, ∠ABC = ∠ACB] 2∠ABC = 180° – 60° 2∠ABC = 120° ∠ABC = 60° Therefore, ∠ABC = ∠ACB = 60° 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°.

Answer»

Given,

In ΔABC

∠ ABC = ∠ ACB

Divide both sides by 2, we get

\(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\)∠ACB

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

Now,

∠BOC = 90° + \(\frac{1}{2}\)∠A

120° – 90° = \(\frac{1}{2}\)∠A

30° x 2 = ∠A

∠A = 60°

Now in ΔABC

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

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

2∠ABC = 180° – 60°

2∠ABC = 120°

∠ABC = 60°

Therefore,

∠ABC = ∠ACB = 60°

Hence, proved