Question

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

In the figure below, segment DE is parallel to segment BC and segment EF is parallel to AB:

The figure shows triangle ABC with segments DE and DF. Point D is on side AB, point E is on side AC, and point F is on side BC. Segment AD is 16, segment AE is 24, segment EC is 36, and segment FC is 27.

Which statement can be proved true using the given theorem?

Segment BD = 32
Segment BD = 36
Segment BF = 15
Segment BF = 18

Answer 1

Using the theorem that a line parallel to one side of a triangle divides the other two proportionately, we find that segment BF is 18.

Step-by-step explanation:

The theorem mentioned, stating that a line parallel to one side of a triangle divides the other two proportionately, applies to the given problem. Since segment DE is parallel to BC and EF is parallel to AB, we can set up proportions using the given segment lengths. According to the problem, AD is 16, AE is 24, and EC is 36. The ratio of AD to AE (16:24) can be simplified to 2:3. Therefore, since DE is parallel to BC and the segments are divided proportionately, BD:DF should be in the same ratio, which is 2:3.

Next, by the same theorem, segment EF is parallel to AB, which makes triangles DEF and ABC similar. Therefore, the ratio of DF to FC should also be 2:3. Given that FC is 27, we can find the length of DF using this proportion: DF/27 = 2/3, which gives us DF = 18. Hence, segment BF, which is the same as DF in this case, equals 18.