Correct option is (C) CE/ED

In \(\triangle BCD,\) EF || DB

\(\therefore\) \(\triangle CEF\sim\triangle CDB\)    (Corresponding angles are equal as EF ||DB)

\(\therefore\) \(\frac{CE}{CD}=\frac{CF}{CB}\)                 (By properties of similar triangles)

\(\Rightarrow\) \(\frac{CE}{CD-CE}=\frac{CF}{CB-CF}\)    \((If\,\frac ab=\frac cd\,then\,\frac a{b-a}=\frac c{d-c})\)

\(\Rightarrow\) \(\frac{CE}{ED}=\frac{CF}{FB}\)

Hence, \(\frac{CF}{FB}\) = \(\frac{CE}{ED}\)

Correct option is: (C) \(\frac{CE}{ED}\)