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Question
In a ΔABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
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Solution
Given,
In ΔABC
∠ABC=∠ ACB
Divide both sides by '2'
`1/2∠ABC=1/2∠ACB`
⇒ ∠OBC=∠ OCB [∵ OB, OC bisects ∠B and ∠C]
Now
`∠BOC=90^@+1/2∠A`
`120^@-90^@=1/2∠A`
`30^@xx(2)=∠A`
`∠A=60^@`
Now in Δ ABC
`∠A+∠ABC+∠ACB=180^@` (Sum of all angles of a triangle)
[∵∠ABC=∠ACB]
`60^@+2∠ABC=180^@`
`2∠ABC=180^@-60^2`
`∠ABC=120^@/2=90^@`
`∠ABC=∠ ACB`
`∴ ∠ACB=60^@`
Hence proved.
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