Video Transcript
In this video, we will learn how to
use the triangle midsegment theorem to prove that lines are parallel or find a
missing side length in a particular triangle scenario.
Let’s begin by looking at the first
of the triangle midsegment theorems. This theorem states that the line
segment passing through the midpoint of one side of a triangle that is also parallel
to another side of a triangle bisects the third side of the triangle. To see this on a diagram, let’s
take this triangle 𝐴𝐶𝐷. Taking the midpoint of one of the
sides, so using the midpoint 𝐵 of line segment 𝐴𝐷, and then drawing a line
parallel to the base, which we can take as the line segment 𝐷𝐶, we would have a
line like this, line segment 𝐵𝐸.
Now, this theorem tells us that
this line segment 𝐵𝐸, which is from the midpoint of one of the sides of the
triangle and parallel to another side, is then a bisector of the third side. Equivalently, that means that this
point 𝐸 must be the midpoint of line segment 𝐴𝐶. Let’s now see how this theorem can
be proved.
Let’s begin with the same triangle
𝐴𝐶𝐷, and we have the line segment 𝐵𝐸 drawn from the midpoint 𝐵, which is
parallel to line segment 𝐶𝐷. As we are trying to prove the
theorem, then we don’t yet know anything about the third side of line segment
𝐴𝐶. We can construct a line 𝐴𝑦 such
that 𝐴𝑦 is parallel to the line segments 𝐵𝐸 and 𝐷𝐶. The reason why we do this will be
clear soon. If we consider line segments 𝐴𝐷
and 𝐴𝐶, these are both transversals of the three parallel lines.
Recall that if a set of parallel
lines divides a transversal into segments of equal lengths, then that set divides
any other transversal into segments of equal lengths. So, because line segment 𝐴𝐷 was
split into two congruent pieces, we know that line segment 𝐴𝐶 will be the
same. It will also be split into two
congruent pieces. In other words, this third side of
the triangle has been bisected, thus proving this triangle midsegment theorem.
Now, the converse of this theorem
also holds true. We can state this as the line
segment joining the midpoints of two sides of a triangle is parallel to its third
side. So, this time, if we take triangle
𝐴𝐶𝐷 and join the midpoints 𝐵 and 𝐸 of line segments 𝐴𝐷 and 𝐴𝐶,
respectively, then, by the converse of the first triangle midsegment theorem, line
segments 𝐵𝐸 and 𝐷𝐶 are parallel. We will now see the final triangle
midsegment theorem. The second part of the triangle
midsegment theorem is all about the lengths of two important line segments.
Before we state this theorem, let’s
do some investigations to prove the result. We can take this triangle 𝑃𝑄𝑅
and draw the line segment 𝑆𝑇, where 𝑆 is the midpoint of line segment 𝑃𝑅 and 𝑇
is the midpoint of line segment 𝑃𝑄. We will try to work out a
relationship between the lengths of line segments 𝑆𝑇 and 𝑅𝑄. To do this, let’s add the ray 𝑆𝑦
such that line segment 𝑆𝑦 is parallel to line segment 𝑃𝑄. Now, we’ve already seen the
property that the line segment joining the midpoints of two sides of a triangle is
parallel to the third side; that was the converse of the first part of the
theorem. Therefore, the line segment 𝑆𝑇 is
parallel to line segment 𝑅𝑄.
Then, because we constructed line
segment 𝑆𝑦 parallel to line segment 𝑃𝑄, we can apply the first triangle
midsegment theorem. The line segment passing through
the midpoint of one side of a triangle that is also parallel to another side of the
triangle bisects the third side of the triangle. 𝑆𝑦 is a line segment from the
midpoint 𝑆 parallel to line segment 𝑃𝑄. And that means that line segment
𝑅𝑄 is bisected. This may a little harder to
visualize at first, but it also illustrates the point that we don’t always need to
have the parallel sides as horizontal with the bisected sides upwards from that. And this means that we now know
that each of the line segments 𝑅𝑦 and 𝑦𝑄 will be equal to half the length of the
whole line segment 𝑅𝑄.
Let’s take this one step further to
see how we can work out the length of the line segment 𝑆𝑇. Notice that within this triangle,
we have the quadrilateral 𝑆𝑇𝑄𝑦. 𝑆𝑇𝑄𝑦 has two pairs of parallel
sides, and so, by definition, it is a parallelogram. Parallelograms have opposite sides
congruent. Therefore, the length of line
segment 𝑆𝑇 will also be one-half 𝑅𝑄. Importantly, that gives us the
relationship that the length of line segment 𝑆𝑇 is half the length of line segment
𝑅𝑄. We can now formally define the
second part of the triangle midsegment theorem. The length of the line segment
joining the midpoints of two sides of a triangle is equal to half the length of the
third side.
Before we look at some questions,
we can note that sometimes what we saw as the converse of part one and part two of
the triangle midsegment theorems are combined into one theorem. This is often referred to
collectively as the triangle midsegment theorem, stated as the line segment
connecting the midpoints of two sides of a triangle is parallel to the third side
and is half its length. We’ll now see a question where we
apply the triangle midsegment theorems to find the perimeter of a shape.
Given that 𝐷 and 𝐸 are the
midpoints of line segments 𝐴𝐵 and 𝐴𝐶, respectively, 𝐴𝐷 equals 32
centimeters, 𝐴𝐸 equals 19 centimeters, and 𝐷𝐸 equals 39 centimeters, find
the perimeter of 𝐷𝐵𝐶𝐸.
We can begin by filling in the
given length information on the figure of 32 centimeters for 𝐴𝐷, 19
centimeters for 𝐴𝐸, and 39 centimeters for 𝐷𝐸. We can also identify the
information that 𝐷 and 𝐸 are midpoints of their respective line segments. So, 𝐸𝐶 also has a length of
19 centimeters, and 𝐷𝐵 has a length of 32 centimeters. We know that we need to work
out the perimeter of this quadrilateral 𝐷𝐵𝐶𝐸. We’ve worked out three of these
four sides, so we still need to work out the length of the line segment
𝐶𝐵. To do this, let’s use the
triangle midsegment theorem.
This theorem states that the
line segment connecting the midpoints of two sides of a triangle is parallel to
the third side and is half its length. This means we can recognize
that line segment 𝐶𝐵 must be parallel to line segment 𝐷𝐸, and we know
something about the length of line segment 𝐶𝐵. Since 𝐷𝐸 is half the length
of 𝐶𝐵, we can also write that 𝐶𝐵 equals two times 𝐷𝐸. We were given that 𝐷𝐸 is 39
centimeters, so doubling this, we can calculate that 𝐶𝐵 is 78 centimeters.
So, we now have enough
information to calculate the perimeter of 𝐷𝐵𝐶𝐸. Recall that the perimeter is
the distance around the outside edge of a shape. So that means we need to add
the four lengths of 39, 32, 78, and 19 centimeters, which gives us the final
answer that the perimeter of 𝐷𝐵𝐶𝐸 is 168 centimeters.
In the next example, we will apply
the triangle midsegment theorem a number of times in the same figure. And when we’re working through a
problem like this, it can be really helpful to highlight the specific line segments
so that we can easily identify the key parts that we’re working with.
In the figure shown, 𝐸, 𝐹,
and 𝐷 are the midpoints of line segments 𝐵𝐶, 𝐴𝐵, and 𝐴𝐶,
respectively. Find the perimeter of triangle
𝐸𝐹𝐷.
We should note that from the
information we are given and the markings on the diagram, that we have three
midpoints of line segments here. 𝐸, 𝐹, and 𝐷 bisect their
respective line segments. So, in order to find the
perimeter of triangle 𝐸𝐹𝐷, that’s the distance around the outside edge, we’ll
need to calculate the lengths of the three line segments 𝐹𝐷, 𝐷𝐸, and
𝐸𝐹.
Now, because we know that there
are some midpoints of lines, that might make us wonder if we could possibly
apply one of the triangle midsegment theorems. We can recall that the length
of the line segment joining the midpoints of two sides of a triangle is equal to
half the length of the third side. Let’s look at line segment
𝐹𝐷. Line segment 𝐹𝐷 is a line
segment joining the midpoints of two sides of a triangle. Therefore, its length is going
to be half the length of the third side, which is line segment 𝐵𝐶. The length of 𝐵𝐶 is given as
4.6 centimeters, so half of this is 2.3 centimeters.
Now, let’s see if we can do the
same to calculate the lengths of the other two sides in triangle 𝐸𝐹𝐷. We can consider the line
segment 𝐷𝐸 next. Line segment 𝐷𝐸 joins the
midpoints of two sides of a triangle, because it joins 𝐷, the midpoint of line
segment 𝐴𝐶, and 𝐸, the midpoint of line segment 𝐵𝐶. Therefore, we know that it must
be half the length of line segment 𝐴𝐵, which is the third side of the
triangle. Half of 5.5 centimeters is 2.75
centimeters. And we can do the same for the
length of line segment 𝐸𝐹. It joins midpoints 𝐸 and
𝐹. So, the length of 𝐸𝐹 will be
half of line segment 𝐴𝐶; half of 6.2 centimeters is 3.1 centimeters.
And now to find the perimeter
of triangle 𝐸𝐹𝐷, we add these three calculated lengths together. 2.3 plus 2.75 plus 3.1 is equal
to 8.15 centimeters. And so, by applying the
triangle midsegment theorem three times, we have determined that the perimeter
of triangle 𝐸𝐹𝐷 is 8.15 centimeters.
Let’s now see how we can apply the
converse of the triangle midsegment theorem to determine an unknown length.
The perimeter of square
𝐴𝐵𝐶𝐷 is 352. Find 𝐴𝐹.
In this question, we need to
find the length of the line segment 𝐴𝐹, which is part of the line segment 𝐴𝐵
on one of the sides of the square. We aren’t given any length
information on the diagram, but we are told that the perimeter of the square is
352 length units.
Now, given that the perimeter
is the distance around the outside edge of a shape, and we know that the shape
is a square, then we can work out one of the side lengths. Because four times the length
of one side would give us the perimeter of a square, to work out the length of
one side of this square, we divide the perimeter of 352 by four. That gives us 88 length
units. So, all the sides of the square
will have a length of 88 length units.
Now, given that we want to find
the length of 𝐴𝐹, we might try to guess at this point, but it’s important to
apply our geometry knowledge in questions like this so that we can prove our
calculated length is correct. So, let’s think about what we
know of the geometry of a square. In particular, we can recall
that in a square, the diagonals bisect one another. So, the diagonals of line
segments 𝐴𝐶 and 𝐷𝐵 are each bisected at 𝑀. But let’s focus on line segment
𝐴𝐶.
By using the triangle 𝐴𝐵𝐶,
we observe that we can apply one of the triangle midsegment theorems here. This theorem states that the
line segment passing through the midpoint of one side of a triangle that is also
parallel to another side of the triangle bisects the third side of the
triangle. Here, we have got the line
segment 𝑀𝐹 passing through the midpoint of one side of a triangle, because we
know that 𝑀 is the midpoint of the line segment 𝐴𝐶. We also know that this same
line segment 𝑀𝐹 is parallel to another side of the triangle because of the
markings on the diagram. And so, that means that the
third side of the triangle must be bisected.
The line segment 𝐴𝐵, which we
calculated has a length of 88 length units, is split into two equal pieces. And 88 divided by two equals
44. And so, by applying this
triangle midsegment theorem, we have found that the length of line segment 𝐴𝐹
is 44 length units.
We can now summarize the key points
of this video. We saw, and proved, the first part
of the triangle midsegment theorem. The line segment passing through
the midpoint of one side of a triangle that is also parallel to another side of the
triangle bisects the third side of the triangle.
The converse of the first theorem
is also true. That is, the line segment joining
the midpoints of two sides of a triangle is parallel to the third side. We also saw that the length of the
line segment joining the midpoints of two sides of a triangle is equal to half the
length of the third side.
The second and third theorems
listed here may often be combined in the theorem that the line segment connecting
the midpoints of two sides of a triangle is parallel to the third side and is half
its length. And so, as displayed in this
diagram, we can see that these theorems give us a relationship between the midpoints
of sides of a triangle and the parallelism and lengths of a midsegment and a side of
the triangle.
