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Question -

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



Answer -

Given : In ∠ABC, BO and CO are the bisectors of ∠B and ∠C respectively and ∠BOC = 120° and ∠ABC = ∠ACB
 
To prove : ∠A = ∠B = ∠C = 60°
Proof : ∵ BO and CO are the bisectors of ∠B and ∠C
∴ ∠BOC = 90° +  ∠A
But ∠BOC = 120°
∴ 90°+   ∠A = 120°
∴ 12 ∠A = 120° – 90° = 30°
∴ ∠A = 60°
∵ ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠B + ∠C = 180° – 60° = 120° and ∠B = ∠C
∵ ∠B = ∠C =  = 60°
Hence ∠A = ∠B = ∠C = 60°

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