Question

Answer 1

a, Ta có: a//b ,  a//c => b//c

b, Do a//b => \(\widehat{A}+\widehat{AOb}=180^o\left(2gócTCP\right)\)

                 =>\(120^o+\widehat{AOb}=180^o\)

                 => \(\widehat{AOb}=60^o\left(1\right)\)

Do b//c => \(\widehat{bOB}+\widehat{OBc}=180^o\left(2gócTCP\right)\)

             => \(\widehat{bOB}+120^o=180^o\)

             => \(\widehat{bOB}=60^o\left(2\right)\)

Từ (1) và (2)=> \(\widehat{AOb}=\widehat{bOB}\)hay Ob là tia pg cuả \(\widehat{AOB}\left(dpcm\right)\)