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In A B C ,\ A B=A C , and the bisectors...

In ` A B C ,\ A B=A C ,` and the bisectors of angles `B\ a n d\ C` intersect at point `Odot` Prove that `B O=C O\ ` and the ray `A O` is the bisector of angle `B A C`
Since the angles opposite to equal sides are equal,
`AB=AC`
`=>/_C=/_B`
`=>2/_B=2/_C`
Since BO and CO are bisectors of `/_B and /_C`, we also have
`/_ABO=2/_B`
`/_ACO= 2/_C`
`/_ABO=2/_B=2/_C ​=/_ACO.`
Consider `/_\ BCO:/_OBC=/_OCB`
`BO=CO`(Sides opposite to equal angles are equal)
Finally, consider triangles ABO and ACO.
`BA=CA`(given)
`BO=CO`(proved)
`/_ABO=/_ACO` (proved)
Hence, by S.A.S postulate`/_\ABO~=/_\ACO`
`/_BAO=/_CAO`
`AO bisects /_A.`

Text Solution

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Since the angles opposite to equal sides are equal,
`AB=AC`
`=>/_C=/_B`
`=>2/_B=2/_C`
Since BO and CO are bisectors of `/_B and /_C`, we also have
`/_ABO=2/_B`
`/_ACO= 2/_C`
`/_ABO=2/_B=2/_C ​=/_ACO.`
Consider `/_\ BCO:/_OBC=/_OCB`
`BO=CO`(Sides opposite to equal angles are equal)
Finally, consider triangles ABO and ACO.
`BA=CA`(given)
`BO=CO`(proved)
`/_ABO=/_ACO` (proved)
Hence, by S.A.S postulate`/_\ABO~=/_\ACO`
`/_BAO=/_CAO`
`AO bisects /_A.`
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