In the figure CD = DA, and EF || BD, CF then CF/FB = …….(A) CD/DA

1.	In the figure CD = DA, and EF \|\| BD, CF then CF/FB = …….(A) CD/DA(B) CE/DA(C) CE/ED(D) CD/CA
Answer» 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) \(\frac{CE}{ED}\)

In the figure CD = DA, and EF || BD, CF then CF/FB = …….(A) CD/DA(B) CE/DA(C) CE/ED(D) CD/CA

Answer»

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) \(\frac{CE}{ED}\)