1. In $∆$OAB, E is the mid point of AB and F is a point on OA such that OF = 2FA. If C is the point of intersection of OE and BF, then find the ratios OC : CE and BC : CF are

$$\begin{gathered} \overline{\mathrm{OA}}=\overline{\mathrm{a}} \overline{\mathrm{OB}}=\overline{\mathrm{b}} \\ \overline{\mathrm{OE}}=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2} \\ \overline{\mathrm{OF}}=\frac{2\overline{\mathrm{a}}}{3} \\ \overline{\mathrm{OC}}=\frac{\mathrm{\lambda }\left({\displaystyle \frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}}\right)+\overline{\mathrm{O}}}{\mathrm{\lambda }+1} \overline{\mathrm{OC}}=\frac{\mathrm{M}\left({\displaystyle \frac{2\overline{\mathrm{a}}}{3}}\right)+5}{\mathrm{M}+1} \\ \mathrm{compare} \mathrm{coff} \Rightarrow \frac{\mathrm{\lambda }}{2(\mathrm{\lambda }+1)}=\frac{2\mathrm{M}}{3(\mathrm{M}+1)}, \frac{\mathrm{\lambda }}{2(\mathrm{\lambda }+1)}=\frac{1}{\mathrm{M}+1} \\ \Rightarrow \frac{2\mathrm{M}}{3}= \Rightarrow \mathrm{M}=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \frac{\mathrm{\lambda }}{2(\mathrm{\lambda }+1)}=\frac{2}{5} \\ 5\mathrm{\lambda }=4\mathrm{\lambda }+4 \Rightarrow \mathrm{\lambda }=4 \\ \frac{\mathrm{OC}}{\mathrm{CE}} =\frac{4}{1} \frac{\mathrm{BC}}{\mathrm{CF}}=\frac{3}{2} \end{gathered}$$