$\textit{Đáp án + Giải thích các bước giải:}$
a) Có tia CO cắt AB tại D
Xét ΔBOD: $\widehat{BOC}$ = $\widehat{BDO}$ + $\widehat{DBO}$
Xét ΔADC: $\widehat{BDO}$ = $\widehat{BAC}$ + $\widehat{ACO}$
⇒ $\widehat{BOC}$ = $\widehat{ABO}$ + $\widehat{BAC}$ + $\widehat{ACO}$ (đpcm)
b) $\widehat{ABO}$ + $\widehat{ACO}$ = 90° - $\dfrac{\widehat{BAC}}{2}$
Có $\widehat{BOC}$ = $\widehat{ABO}$ + $\widehat{BAC}$ + $\widehat{ACO}$
$\widehat{BOC}$ - $\widehat{BAC}$ = 90° - $\dfrac{\widehat{BAC}}{2}$}$
$\widehat{BOC}$ = 90° + $\dfrac{\widehat{BAC}}{2}$
$\widehat{DCB}$ = 180° - $\widehat{BOC}$ - $\widehat{CBO}$
⇒ $\widehat{DCB}$ = 180° - 90° - $\bigg(\dfrac{\widehat{BAC}}{2} + \dfrac{ABC}{2}\bigg)$
⇒ $\widehat{DCB}$ = 90° - $\dfrac{\widehat{BAC}$ - \widehat{ABC}}{2}$
⇒ $\widehat{DCB}$ = $\dfrac{180°}{2}$ - $\dfrac{\widehat{BAC} - \widehat{ABC}}{2}$
⇒ $\widehat{DCB}$ = $\dfrac{180° - \widehat{BAC} - \widehat{ABC}}{2}$
Có $\widehat{BAC}$ + $\widehat{ABC}$ + $\widehat{ACB}$ = 180° (Định lí tổng 3 góc trong cùng Δ)
⇒ $\widehat{DCB}$ = $\dfrac{\widehat{ACB}}{2}$
⇒ CO là tia phân giác của $\widehat{ABC}$ (đpcm)
