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In a triangle A B C ,/A B C=/A C B and t...

In a triangle `A B C ,/_A B C=/_A C B` and the bisectors of `/_A B C` and `/_A C B` intersect at `O` such that `/_B O C=120^@dot` Show that `/_A=/_B=/_C=60^@`

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in `/_\OBC`
`/_BOC=120^@`
by angle sum ,`/_OBC+/_OCB+120^@=180^@`
`2/_OBC=60^@` [since `/_B=/_C` so `1/2/_B=1/2/_C`
`/_OBC =/_OCB`]
`/_OBC=/_OCB=30^@`
so `/_B=/_C=60^@`
by angle sum of ABC
`/_A=180^@-60^@-60^@`
`/_A=60^@`
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