Question

Theorem: A line parallel to one side of a triangle divides the other two proportionately.

Answer 1

Answer:

Segment BF = 16 is true.

Step-by-step explanation:

Since, DE is parallel to BC, so DE will divide AB and AC proportionally.

Hence,

$\frac{AE}{EC} = \frac{AD}{DB}$

⇒  $\frac{12}{18} = \frac{6}{DB}$ 

{Since, given that AE = 12, EC = 18 and AD = 6}

⇒ BD = 9.

Again, since, EF is parallel to AB, so EF will divide BC and AC proportionally.

Hence, $\frac{CE}{AE} = \frac{FC}{BF}$

⇒ $\frac{18}{12} = \frac{24}{BF}$   

{Since, given that AE = 12, EC = 18 and FC= 24}

⇒ BF = 16.

Therefore, segment BF = 16 is true. (Answer)