Bisector `A D` of `/_B A C` of ` A B C` passes through the centre `O` of the circumcircle of ` A B C` as shown in figure. Prove that `A B=A Cdot`

Explanation

Given: Bisector `AD` of `/_BAC` of `/_\ABC` passed through the centre `O` of the circumcircle of `/_\ABC`
To Prove: `AB=AC`
Construction: Draw `OP_|_AB` and `OQ_|_AC`
Proof: In `/_\APO` and `/_\AQO`,
`/_OPA=/_OQA` [Each`=90^@` (By construction)]
`/_OAP=/_OAQ` [Given]
`OA=OA` [Common]
`:./_\APO~=/_\AQO` [By `SAS`]
`:. OP=OQ` [By `CPCT`]
`:. AB=AC` [since, Chords equidistant from the centre are equal]
Hence, `AB=AC`