Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\Delta AOB;C \in AO;D \in BO;CD//AB\\
\Rightarrow \dfrac{{CO}}{{AO}} = \dfrac{{DO}}{{BO}} = \dfrac{{CD}}{{AB}}\left( {DL:Thales} \right)\\
\Rightarrow \left\{ \begin{array}{l}
CO = \dfrac{{AO.DO}}{{BO}} = \dfrac{{12.3}}{9} = 4cm\\
CD = \dfrac{{AB.DO}}{{BO}} = \dfrac{{18.3}}{9} = 6cm
\end{array} \right.
\end{array}$$\begin{array}{l}
\Delta AOB;C \in AO;D \in BO;CD//AB\\
\Rightarrow \dfrac{{CO}}{{AO}} = \dfrac{{DO}}{{BO}} = \dfrac{{CD}}{{AB}}\left( {DL:Thales} \right)\\
\Rightarrow \left\{ \begin{array}{l}
CO = \dfrac{{AO.DO}}{{BO}} = \dfrac{{12.3}}{9} = 4cm\\
CD = \dfrac{{AB.DO}}{{BO}} = \dfrac{{18.3}}{9} = 6cm
\end{array} \right.
\end{array}$
Vậy $OC=4cm;CD=6cm$
b) Ta có:
$\begin{array}{l}
\Delta FAB;C \in FB;D \in FA;CD//AB\\
\Rightarrow \dfrac{{FA}}{{AD}} = \dfrac{{FC}}{{CB}}\left( {DL:Thales} \right)\\
\Rightarrow FA.BC = FC.AD
\end{array}$
c) Ta có:
$\begin{array}{l}
\Delta ABD;O \in BD;M \in AD;OM//AB\\
\Rightarrow \dfrac{{OM}}{{AB}} = \dfrac{{DO}}{{DB}}\left( 1 \right)
\end{array}$
Lại có:
$\begin{array}{l}
\Delta ABC;O \in AC;N \in BC;ON//CD\\
\Rightarrow \dfrac{{ON}}{{AB}} = \dfrac{{CO}}{{CA}}\left( 2 \right)
\end{array}$
Mặt khác:
$\begin{array}{l}
\Delta AOB;C \in AO;D \in BO;CD//AB\\
\Rightarrow \dfrac{{BO}}{{DO}} = \dfrac{{AO}}{{CO}}\\
\Rightarrow \dfrac{{BO + DO}}{{DO}} = \dfrac{{AO + CO}}{{CO}}\\
\Rightarrow \dfrac{{BD}}{{DO}} = \dfrac{{AC}}{{CO}}\\
\Rightarrow \dfrac{{DO}}{{DB}} = \dfrac{{CO}}{{CA}}\left( 3 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right),\left( 3 \right) \Rightarrow \dfrac{{OM}}{{AB}} = \dfrac{{ON}}{{AB}} \Rightarrow OM = ON$
