Correct option is (D) BF/FC
Given AB || DC & EF || AB
\(\therefore\) DC || EF
Construction :
Join BD which intersect EF at G.
In \(\triangle ABD,\) EG || AB
\(\therefore\) \(\triangle EDG\sim\triangle ADB\) (Corresponding angles are equal as EG || AB)
\(\therefore\) \(\frac{ED}{AD}=\frac{GD}{BD}\) (Properties of similar triangles)
\(\therefore\) \(\frac{ED}{AD-ED}=\frac{GD}{BD-GD}\) \((\because\,If\,\frac ab=\frac cd\Rightarrow\frac a{b-a}=\frac c{d-c})\)
\(\Rightarrow\) \(\frac{ED}{AE}=\frac{GD}{BG}\)
\(\Rightarrow\) \(\frac{AE}{ED}=\frac{BG}{GD}\) ________________(1)
In \(\triangle BCD,\) GF || DC
\(\therefore\) \(\triangle BGF\sim\triangle BDC\)
\(\therefore\) \(\frac{BG}{BD}=\frac{BF}{BC}\)
\(\Rightarrow\) \(\frac{BG}{BD-BG}=\frac{BF}{BC-BF}\)
\(\Rightarrow\) \(\frac{BG}{GD}=\frac{BF}{FC}\) ________________(2)
From (1) & (2), we obtain
\(\frac{AE}{ED}\) = \(\frac{BF}{FC}\)