Giải thích các bước giải:
Kẻ $BD//OC, D\in OA$
Vì $OC$ là phân giác $\widehat{AOB}\to \widehat{BOC}=\widehat{COA}=\dfrac12\widehat{AOB}=60^o$
Mà $BD//OC\to \widehat{ODB}=\widehat{COA}=60^o=\widehat{BOC}=\widehat{BDO}$
$\to\Delta OBD$ đều
$\to OB=OD=BD$
Ta có $OC//BD$
$\to \dfrac{AO}{AD}=\dfrac{OC}{BD}$
$\to \dfrac{AD}{AO}=\dfrac{BD}{OC}$
$\to \dfrac{OD+OA}{AO}=\dfrac{BD}{OC}$
$\to \dfrac{OD}{OA}+1=\dfrac{BD}{OC}$
$\to \dfrac{OB}{OA}+1=\dfrac{OB}{OC}$
$\to \dfrac{1}{OA}+\dfrac{1}{OB}=\dfrac{1}{CO}$
