In a triangle ABC, 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, CF = 3 cm, then find the length (in cm) of EF.

2

In \(\Delta \)ABC , DF is parallel to AC,

= \(\frac{BD}{AB}\) = \(\frac{BF}{BC}\)    ..(1)

In \(\Delta \)BAF, DE is parallel to AF,

= \(\frac{BE}{BF}\) = \(\frac{BD}{AB}\)     ..(2)

From equation 1 and 2,

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

= 4(4 + 3 + EF) = (4 + EF)²  

= 28 + 4EF = 16 + 8EF + EF²

= EF² + 6EF - 2EF - 12 = 0

= (EF - 2)(3F + 6) = 0

= (EF - 2)(EF + 6) = 0

EF = 2 or EF = -6

Therefore, length of EF is 2 cm.