Lesson Video: Triangle Midsegment Theorems Mathematics

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.