Lesson Explainer: Triangle Midsegment Theorems | Nagwa Lesson Explainer: Triangle Midsegment Theorems | Nagwa

Lesson Explainer: Triangle Midsegment Theorems Mathematics

In this explainer, we will learn how to use the triangle midsegment theorem to prove the parallelism of lines in a triangle or find a missing side length.

Let’s begin with understanding what the triangle midsegment theorem states.

Theorem: Triangle Midsegment Theorem (Part 1)

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.

We can prove this by considering triangle 𝐴𝐶𝐷 with line segment 𝐸𝐵 that passes through the midpoint, 𝐸, of 𝐴𝐷 and that is parallel to 𝐷𝐶.

We can construct line 𝐴𝑌 such that 𝐴𝑌𝐸𝐵𝐷𝐶.

Line segments 𝐴𝐷 and 𝐴𝐶 are transversals of these three parallel lines. We 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.

Given that 𝐴𝐷 was split into two congruent segments by the parallel lines, 𝐴𝑌, 𝐸𝐵, and 𝐷𝐶, then the other transversal, 𝐴𝐶, must also be split into two congruent segments.

Hence, 𝐴𝐵=𝐵𝐶.

Therefore, we have proven that the third side of the triangle has been bisected.

The converse of this theorem is also true, that is, if we have a triangle where two sides are bisected by a line segment, then that line segment is parallel to the third side. This is defined below.

Theorem: Converse of the Triangle Midsegment Theorem (Part 1)

The line segment joining the midpoints of two sides of a triangle is parallel to the third side.

We can also see one more of the triangle midsegment theorems.

Theorem: Triangle Midsegment Theorem (Part 2)

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 examine how we can prove this theorem. Consider 𝐴𝐶𝐷, where 𝐸 and 𝐵 are the midpoints of 𝐴𝐷 and 𝐴𝐶, respectively, such that 𝐴𝐸=𝐸𝐷 and 𝐴𝐵=𝐵𝐶. We can construct ray 𝐸𝐹 such that 𝐸𝐹𝐴𝐶 and 𝐸𝐹 intersects 𝐷𝐶 at 𝐹.

We can then apply the triangle midsegment theorem, which states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side. This means that 𝐸𝐵𝐷𝐶.

As we constructed 𝐸𝐹𝐴𝐶, then by the triangle midsegment theorem, which 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, we have that 𝐷𝐶 is bisected. Hence, 𝐷𝐹=𝐹𝐶, and 𝐹 is the midpoint of 𝐷𝐶. Furthermore, 𝐹𝐶=12𝐷𝐶.

We can then consider quadrilateral 𝐸𝐵𝐶𝐹.

𝐸𝐵𝐶𝐹 is a quadrilateral with two pairs of opposite sides parallel, which by definition is a parallelogram. In a parallelogram, opposite sides are congruent. Hence, 𝐸𝐵=𝐹𝐶=12𝐷𝐶.

Therefore, we have proven the theorem: the line segment joining the midpoints of two sides of a triangle is equal to half the length of the third side.

It is worth noting that theorems 1 and 2 here are often referred to collectively as the triangle midsegment theorem, stated as follows: the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half its length.

In the following questions, we will see how we can apply the triangle midsegment theorem and its converse in order to find unknown side lengths, beginning with a question where we need to find the perimeter of a shape.

Example 1: Finding the Perimeter of a Quadrilateral Using the Triangle Midsegment Theorem

Given that 𝐷 and 𝐸 are the midpoints of 𝐴𝐵 and 𝐴𝐶, respectively, 𝐴𝐷=32cm, 𝐴𝐸=19cm, and 𝐷𝐸=39cm, determine the perimeter of 𝐷𝐵𝐶𝐸.

Answer

We can begin by filling in the given length information on the figure, which is that 𝐴𝐷=32cm, 𝐴𝐸=19cm, and 𝐷𝐸=39cm.

As 𝐷 and 𝐸 are the midpoints of 𝐴𝐵 and 𝐴𝐶, we know that 𝐸𝐶=𝐴𝐸=19cm and 𝐷𝐵=𝐴𝐷=32.cm

The triangle midsegment 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.

Thus, we can say that 𝐶𝐵𝐷𝐸 and 𝐶𝐵=2×(𝐷𝐸).

Given that 𝐷𝐸=39cm, we have 𝐶𝐵=2×39=78.cm

Finally, we need to calculate the perimeter of 𝐷𝐵𝐶𝐸. This is the distance around the outside of 𝐷𝐵𝐶𝐸, the trapezoid on the lower part of the figure. Using 𝐷𝐸=39cm, 𝐸𝐶=19cm, 𝐶𝐵=78cm, and 𝐷𝐵=32cm, we have perimeterofcm𝐷𝐵𝐶𝐸=𝐷𝐸+𝐸𝐶+𝐶𝐵+𝐷𝐵=39+19+78+32=168.

Thus we have the answer that the perimeter of 𝐷𝐵𝐶𝐸 is 168 cm.

In the next example, we will apply the triangle midsegment theorem a number of times in different triangles in the same figure. When working through a problem such as this, it can be helpful to outline or highlight the specific triangles so that we can correctly identify the key segments that we are working with.

Example 2: Applying the Triangle Midsegment Theorem to Solve a Problem

In the figure shown, 𝐸, 𝐹, and 𝐷 are the midpoints of 𝐵𝐶, 𝐴𝐵, and 𝐴𝐶 respectively. Find the perimeter of 𝐸𝐹𝐷.

Answer

We are given the information that 𝐸, 𝐹, and 𝐷 are the midpoints of 𝐵𝐶, 𝐴𝐵, and 𝐴𝐶 respectively. In order to calculate the perimeter of 𝐸𝐹𝐷, we will need to determine the lengths of 𝐹𝐷, 𝐷𝐸, and 𝐸𝐹.

To do this, 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 consider 𝐹𝐷.

𝐹𝐷 is a line segment connecting the midpoints of two sides of a triangle. Thus, 𝐹𝐷 is half the length of 𝐵𝐶 (4.6 cm). Therefore, we have 𝐹𝐷=12×4.6=2.3.cm

In the same way, we can consider 𝐷𝐸.

By applying the triangle midsegment theorem again, we have that 𝐷𝐸 must be half the length of 𝐴𝐵 (5.5 cm). Thus, we have 𝐷𝐸=12×5.5=2.75.cm

Finally, we can calculate the length of 𝐸𝐹 in the same way.

𝐸𝐹 must be half the length of 𝐴𝐶 (6.2 cm). Thus, we have 𝐸𝐹=12×6.2=3.1.cm

Hence, as 𝐹𝐷=2.3cm, 𝐷𝐸=2.75cm, and 𝐸𝐹=3.1cm, we can calculate the perimeter of 𝐸𝐹𝐷 as perimeterofcm𝐸𝐹𝐷=𝐹𝐷+𝐷𝐸+𝐸𝐹=2.3+2.75+3.1=8.15.

We can give the answer that the perimeter of 𝐸𝐹𝐷 is 8.15 cm.

In the next example, we will see how we can apply our knowledge of the triangle midsegment theorem to help us prove geometrical properties within a given figure.

Example 3: Completing a Proof Using the Triangle Midsegment Theorem

In the given figure, 𝐸 and 𝐹 are the midpoints of 𝐴𝐵 and 𝐴𝐶, respectively, 𝐵𝐷=12𝐵𝐶, and 𝐵 lies on 𝐷𝐶. What is the shape of 𝐸𝐹𝐵𝐷?

Answer

We are given that 𝐸 and 𝐹 are the midpoints of 𝐴𝐵 and 𝐴𝐶 respectively. Using the triangle midsegment theroem, we know that the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half its length.

Hence, 𝐸𝐹𝐵𝐶 and 𝐸𝐹=12𝐵𝐶.

We are asked to identify the shape of 𝐸𝐹𝐵𝐷.

𝐸𝐹𝐵𝐷 appears to be a parallelogram; however, we must prove this to be the case. A parallelogram is defined as a quadrilateral with two pairs of opposite sides parallel. We have demonstrated by the triangle midsegment theorem that 𝐸𝐹𝐵𝐶, and since we are given that 𝐵 lies on 𝐷𝐶, then 𝐸𝐹𝐵𝐷.

We were given in the question that 𝐵𝐷=12𝐵𝐶, and we proved that 𝐸𝐹=12𝐵𝐶. Hence, 𝐸𝐹 is congruent to 𝐵𝐷.

We have now demonstrated that 𝐸𝐹𝐵𝐷 has a pair of opposite sides, 𝐸𝐹 and 𝐵𝐷, that are parallel and congruent. Hence, 𝐸𝐹𝐵𝐷 must be a parallelogram.

We will now see an example of how we can apply the converse of the triangle midsegment theorem to identify an unknown length.

Example 4: Applying the Triangle Midsegment Theorem to Solve Problems given the Perimeter

The perimeter of square 𝐴𝐵𝐶𝐷 is 352. Find 𝐴𝐹.

Answer

In this question, we are not given any information about the lengths of any line segments. However, we are given the information that the perimeter of this square is 352 length units. Given that the perimeter is the distance around the outside edge and that a square has 4 congruent sides, we can calculate the length of one side as lengthofonesideofthesquarelengthunits=352÷4=88.

At this point, we might try to guess the length of 𝐴𝐹; however, in questions like this, we must apply our geometry knowledge to demonstrate and prove that our calculated length is correct.

We can recall that the diagonals of a square bisect one another, so the diagonals 𝐴𝐶 and 𝐷𝐵 are bisected at 𝑀. Therefore, we have that 𝑀 is the midpoint of 𝐴𝐶.

We can then apply the converse of the triangle midsegment theorem, which 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.

Hence, 𝐴𝐵 is bisected by 𝑀𝐹, and 𝐴𝐹=𝐹𝐵.

We calculated earlier that the length of one side is 88 length units, so 𝐴𝐵=88 length units. Since 𝐴𝐹=𝐹𝐵, then 𝐹 is the midpoint of 𝐴𝐵 and 𝐴𝐹=12𝐴𝐵. We can determine the length of 𝐴𝐹 as 𝐴𝐹=12𝐴𝐵=12×88=44.lengthunits

Therefore, we can give the answer that 𝐴𝐹 is 44 length units.

In the final example, we will see how we can use the converse of the triangle midsegment theorem to prove a geometric property in a given figure.

Example 5: Completing a Proof Using the Triangle Midsegment Theorem

In the given figure, which of the following is true?

  1. 𝐸 is the midpoint of 𝐹𝐺.
  2. 𝐹 is the midpoint of 𝐴𝐷.
  3. 𝐹𝐺=12𝐴𝐵
  4. 𝐶𝐷=12𝐴𝐵

Answer

In the figure, we can observe that we have two pairs of congruent side lengths: 𝐴𝐸 and 𝐸𝐶 and 𝐵𝐺 and 𝐺𝐶. Therefore, we can state that 𝐸 and 𝐺 are the midpoints of 𝐴𝐶 and 𝐵𝐶 respectively. We can recall that by the triangle midsegment theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half its length.

Thus, in the figure, we have that 𝐸𝐺𝐴𝐵𝐶𝐷.

We will now consider the choices that we are presented with and determine which of them is true.

In choice A, we address the statement that 𝐸 is the midpoint of 𝐹𝐺. In fact, we have already proven that 𝐸 is the midpoint of 𝐴𝐶. We must be careful not to confuse the two line segments. Here, we cannot prove that 𝐸 is the midpoint of 𝐹𝐺.

Next, let’s see the statement in choice 𝐵: 𝐹 is the midpoint of 𝐴𝐷. We can consider triangle 𝐴𝐶𝐷.

Given that 𝐹𝐸 passes though the midpoint of 𝐴𝐶 and is parallel to 𝐶𝐷, then by the converse of the triangle midsegment theorem, it must bisect the third side, 𝐴𝐷. Hence, 𝐴𝐹=𝐹𝐷, and 𝐹 is the midpoint of 𝐴𝐷. The statement given in choice B is true.

In choice C, we need to determine if the statement that 𝐹𝐺=12𝐴𝐵 is true. Let’s consider 𝐴𝐵𝐶. Applying the triangle midsegment theorem here would allow us to prove that 𝐸𝐺=12𝐴𝐵.

Since 𝐹 can be observed not to lie on point 𝐸 and 𝐹𝐺𝐸, then 𝐹𝐺 cannot also be the same length as 𝐸𝐺. Thus, this statement is not true.

Finally, let’s consider choice D: 𝐶𝐷=12𝐴𝐵. We cannot apply the triangle midsegment theorem or its converse to demonstrate that this statement is true. As with choice C, the line segment that can be demonstrated to be half the length of 𝐴𝐵 is 𝐸𝐺, not 𝐶𝐷.

Therefore, the true statement is given in choice B: 𝐹 is the midpoint of 𝐴𝐷.

We have seen how we can apply the triangle midsegment theorem and its converse. We can now summarize the key points.

Key Points

  • 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 line segment joining the midpoints of two sides of a triangle is parallel to the third side.
  • 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.

Download the Nagwa Classes App

Attend sessions, chat with your teacher and class, and access class-specific questions. Download the Nagwa Classes app today!

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy