In a triangle ABC, a point D lies on AB and points E and F lie on BC such that DF is parallel to AC and DE is parallel to AF. If BE = 4 cm, EF = 6 cm, then find the length (in cm) of BC.

25

In \(\Delta \)BED and \(\Delta \)BFA

\(\frac{BE}{BF}\) = \(\frac{ED}{FA}\) = \(\frac{BD}{BA}\)

In \(\Delta \)BFD and \(\Delta \)BCA

\(\frac{BF}{BC}\) = \(\frac{FD}{CA}\) = \(\frac{BD}{BA}\)

Now,

\(\frac{BE}{BF}\) = \(\frac{BF}{BC}\)

= \( {BF }^{2 } \) = BE x BC

= \( {(BE \; + \; EF)  }^{2 } \) =  BE x BC

= \( {(4 \; + \; 6)  }^{2 } \) =  4 x BC

= \( {10 }^{2 } \) = 4BC

= BC = 25 cm

Therefore, BC is 25 cm.