Đáp án:
1)
$\begin{array}{l}
OA.OB = O{C^2}\\
\Rightarrow \dfrac{{OA}}{{OC}} = \dfrac{{OC}}{{OB}}\\
Xét:\Delta OAC;\Delta OCB:\\
+ \dfrac{{OA}}{{OC}} = \dfrac{{OC}}{{OB}}\\
+ \widehat O\,chung\\
\Rightarrow \Delta OAC \sim \Delta OCB\left( {c - g - c} \right)\\
2)\Delta OAC \sim \Delta OCB\\
\Rightarrow \left\{ \begin{array}{l}
\widehat {OAC} = \widehat {OCB}\\
\widehat {OCA} = \widehat {OBC}
\end{array} \right.
\end{array}$