In ΔABC, DE || AB, where D and E are points on sides AC and BC, respectively. F is a point between C and D such that EF || BD. If AD = 15 cm, DC = 10 cm, then length of CF is :

4 cm

It is given that :-

AD = 15 cm and DC = 10 cm

In triangle ABC and CDE

\(\frac{CD}{AD}\) = \(\frac{CE}{CB}\) = \(\frac{CD}{CD + AD}\)

\(\frac{CD}{AD}\) = \(\frac{CE}{CB}\) = \(\frac{10}{10 + 15}\)

\(\frac{CD}{AD}\) = \(\frac{CE}{CB}\) = \(\frac{10}{25}\)

\(\frac{CD}{AD}\) = \(\frac{CE}{CB}\) = \(\frac{2}{5}\)

In triangle CFE and CDB

\(\frac{CF}{CD}\) = \(\frac{CE}{CB}\)

{ We know, \(\frac{CE}{CB}\)  = \(\frac{2}{5}\) }

\(\frac{CF}{CD}\) = \(\frac{2}{5}\)

\(\frac{CF}{10}\) = \(\frac{2}{5}\)

CF = 4 cm