tg AOB : BC // AD => tg BOC ~ tg AOD
<=> \(\dfrac{OB}{OA}=\dfrac{OC}{OD}\Leftrightarrow\dfrac{OB}{OB+BA}=\dfrac{OC}{OC+CD}\Leftrightarrow\dfrac{4}{9}=\dfrac{OC}{OC+15}\)
\(\Leftrightarrow OC=\dfrac{4}{9}\left(OC+15\right)\Leftrightarrow\dfrac{5}{9}OC=\dfrac{20}{3}\Leftrightarrow OC=12\)
Vậy OC =12(cm)