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.
Answer & explanation
Correct answer: option 3
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.