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 EF. Point D is on side AB, point E is on side AC and point F is on side BC.

Which statement can be proved true using the given theorem? (6 points)

Select one:

a. Segment BF = 16

b. Segment BD = 20 Incorrect

c. Segment BD = 15

d. Segment BF = 32

C

Step-by-step explanation:

Since DE is parallel to BC then the 2 sides are divided into the same proportion, that is

= , substitute values

= = ( cross- multiply )

6BD = 90 ( divide both sides by 6 )

BD = 15 → C

=====================================

explanation

the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.

the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.

the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers

the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)

m = slope

m = (y2-y1)/(x2-x1)

m = (10-1)/(5-2)

m = 9/3

m = 3 so we get a slope of 3

choice d is the other answer