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.
Answer & explanation
Correct answer: option 3
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.