Question

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

Answer 1

Answer:

Segment BF = 16

Step-by-step explanation:

The given theorem states that a line parallel to one side of a triangle divides the other two sides proportionately

The given theorem is the Triangle Proportionality Theorem

According to the theorem, given that segment DE is parallel to segment BC, we have;

$\dfrac{AD}{BD} = \dfrac{AE}{EC}$

Therefore;

$BD = \dfrac{AD}{\left(\dfrac{AE}{EC} \right) } = AD \times \dfrac{EC}{AE}$

Which gives;

$BD = 6 \times \dfrac{18}{12}= 9$

Similarly, given that EF is parallel to AB, we get;

$\dfrac{AE}{EC} = \dfrac{BF}{FC}$

Therefore;

$BF = FC \times \dfrac{AE}{EC}$

Which gives;

$BF = 24 \times \dfrac{12}{18} = 16$

Therefore, the statement that can be proved using the given theorem is segment BF = 16.