In $\triangle$ ABC, D and E are the points on sides AC and BC, respectively such that DE parallel to AB. F is a point on CE such that DF $\perp$ CE. If CE = 6 cm, and CF = 2.5 cm, then BC is equal to:

14.4 cm

In triangle ABC ,

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

\(\frac{6}{CB}\) = \(\frac{CD}{CA}\)    ----(1)

In triangle AEC,

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

\(\frac{2.5}{6}\) = \(\frac{CD}{CA}\)

\(\frac{5}{12}\) = \(\frac{CD}{CA}\)

By using equation 1 ,

\(\frac{6}{CB}\) = \(\frac{5}{12}\) 

BC = 14.4 cm