Lesson Explainer: Medians of Triangles | Nagwa Lesson Explainer: Medians of Triangles | Nagwa

Lesson Explainer: Medians of Triangles Mathematics • Second Year of Preparatory School

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In this explainer, we will learn how to identify the medians of a triangle and use their properties of proportionality to find a missing length.

The medians of triangles are special lines with special properties. Let us start with defining a median.

Definition: Median

The medians of a triangle are the three segments going from each vertex to the midpoint of the opposite side.

The following diagram shows an example containing a median of a triangle and two other lines in the triangle that are not medians.

As there are 3 vertices in a triangle, there are 3 medians. When we draw all of them, we observe that they all intersect. This is a general property of medians that we will outline in the theorem below.

Theorem: Concurrence of Medians of a Triangle

The three medians of a triangle intersect at a single point (i.e., they are concurrent). Their point of intersection is called the point of concurrence of the medians.

Another property of the medians of a triangle is that the relative position of the point of concurrence of the medians is always the same. Indeed, the point of concurrence is located at two-thirds of the length of the median from the vertex.

Let us visualize this. We can split each median into three-thirds as shown in the diagram. Two-thirds are between the vertex and the point of concurrence, and one-third is between the point of concurrence and the midpoint of the side.

Equivalently, it means that the length of the segment between the vertex and the point of concurrence is twice that between the point of concurrence and the midpoint of the opposite side.

This can be found using algebra as well.

If 𝐴𝑃=23⋅𝐴𝐸 and 𝑃𝐸=13⋅𝐴𝐸, then, from the second equation (by multiplying both sides by 3), we get 𝐴𝐸=3𝑃𝐸, and by substituting 𝐴𝐸 with 3𝑃𝐸 into the first equation, we get 𝐴𝑃=23β‹…3𝑃𝐸, which gives 𝐴𝑃=2𝑃𝐸.

Let us recap this.

Theorem: Position of the Point of Concurrence of the Medians of a Triangle

The distance from each vertex of a triangle to the point of concurrence of its medians is two-thirds of the length of the median from this vertex.

Equivalently, the distance from the point of concurrence of the medians to a vertex is twice that of the distance to the opposite midpoint.

Let us look at our first example, where we need to use the property given in this theorem on the position of the point of concurrence of the medians of a triangle.

Example 1: The Properties of the Point of Concurrence of the Medians of a Triangle

In a triangle 𝐴𝐡𝐢, 𝑀 is the point of concurrency of its medians. If 𝐴𝐷 is a median, then 𝐴𝑀=𝑀𝐷.

Answer

Recall that a median of a triangle is the line segment that joins a vertex with the midpoint of the opposite side. Let us sketch all medians of a triangle 𝐴𝐡𝐢, with 𝑀 as their point of concurrence. 𝐴𝐷 is the median that joins vertex 𝐴 with 𝐷; hence, 𝐷 is the midpoint of side 𝐡𝐢.

We know from the theorem on the point of concurrence of the three medians of a triangle that the distance from each vertex to the point of concurrence of the medians is two-thirds of the length of the median from this vertex. It means that we have here 𝐴𝑀=23𝐴𝐷.

It means that if we cut 𝐴𝐷 into three equal segments, 𝐴𝑀 will be made of two of them. It follows that 𝑀𝐷 is made of the third one. Therefore, 𝐴𝑀 is twice as long as 𝑀𝐷.

Hence, 𝐴𝑀=2𝑀𝐷.

In our second example, we need to identify medians and then use the property of the point of concurrence of medians to find the length from a vertex to the point of concurrence.

Example 2: Identifying Medians and Using the Property of their Point of Concurrence to Find a Missing Length

Find the length 𝐴𝑀, given that 𝐴𝐸=54.

Answer

Looking at the diagram, we observe that both 𝐴𝐸 and 𝐢𝐷 are line segments that join a vertex to the midpoint of the opposite side. Hence, 𝐴𝐸 and 𝐢𝐷 are medians of triangle 𝐴𝐡𝐢. Point 𝑀 is thus the point of concurrence of the medians of triangle 𝐴𝐡𝐢.

Recall that the length of 𝐴𝑀, that is, the distance from the vertex to the point of concurrence, is two-thirds that of the median 𝐴𝐸: 𝐴𝑀=23⋅𝐴𝐸=23β‹…54=36.

Thus, the length of 𝐴𝑀 is 36.

Let us now look at an example, where we use our knowledge about medians of a triangle to form and solve a linear equation.

Example 3: Using the Properties of the Point of Concurrence of the Medians of a Triangle to Form and Solve a Linear Equation

In △𝐾𝑀𝐻, 𝐾𝑄=2 and 𝑄𝑃=(5π‘₯βˆ’7). Find π‘₯.

Answer

In triangle 𝐾𝑀𝐻, 𝑄 is the point of concurrence of its medians. 𝐾 is a vertex and 𝑃 is the middle of the opposite side, 𝐻𝑀. Recall that the distance from each vertex of a triangle to the point of concurrence of its medians is two-thirds of the total length of the median from this vertex. Therefore, for the median 𝐾𝑃, we have 𝐾𝑄=23𝐾𝑃 and 𝑄𝑃=13𝐾𝑃.

As 𝐾𝑄=2Γ—ο€Ό13πΎπ‘ƒοˆ, we see that 𝐾𝑄=2𝑄𝑃.

It is said in the question that 𝐾𝑄=2 and 𝑄𝑃=5π‘₯βˆ’7.

Hence, we have 2=2(5π‘₯βˆ’7).

Dividing both sides of this equation by 2 gives 1=5π‘₯βˆ’7.

Adding 7 to both sides gives 8=5π‘₯.

And, finally, dividing both sides by 5, we find that π‘₯=85=1.6.

Let us now look at the medians in a right triangle. Remember that a right triangle is half of a rectangle, as shown in the following diagram, where 𝐴𝐡𝐢𝐷 is a rectangle and △𝐴𝐡𝐢 and △𝐢𝐷𝐴 are congruent right triangles.

In △𝐴𝐡𝐢, 𝐡𝐸 is a median. In rectangle 𝐴𝐡𝐢𝐷, 𝐡𝐸 is half of a diagonal. As the diagonals of rectangles bisect each other (this is a property of parallelograms) and are equal in length (this is a property of rectangles; it comes from the fact that △𝐴𝐡𝐢 and △𝐢𝐷𝐴 are congruent), we have 𝐡𝐷=𝐴𝐢, so 12𝐡𝐷=12𝐴𝐢, that is, 𝐡𝐸=𝐴𝐸=𝐸𝐢.

We have proven the following property.

Theorem: Length of the Median in a Right Triangle

In a right triangle, the length of the median drawn from the vertex of the right angle equals half the length of the triangle’s hypotenuse.

It is worth noting that one implication of this theorem is that the median drawn from the vertex of the right angle always splits the right triangle into two isosceles triangles.

Let us use this last theorem in our next example and discover at the same time its implication in a special right triangle, namely, the 30∘-60∘ right triangle.

Example 4: Finding the Length of the Smaller Side in a 30Β°-60Β° Right Triangle Using the Properties of Its Median Drawn from the Right Angle

Determine the lengths of 𝐡𝐷 and 𝐴𝐡.

Answer

From the diagram, we observe that △𝐴𝐡𝐢 is a right triangle at 𝐡, and as 𝐷 is the midpoint of 𝐴𝐢, 𝐡𝐷 is the median of △𝐴𝐡𝐢 drawn from its right angle.

Recall that in a right triangle the length of the median drawn from the vertex of the right angle equals half the length of the triangle’s hypotenuse. Hence, we have 𝐡𝐷=12𝐴𝐢=12Γ—49=24.5.cm

We are asked as well to find the length of 𝐴𝐡. The measure of the angle at vertex 𝐢 is given in the diagram: it is 30∘. As the angle at 𝐡 is a right angle, it means that the measure of the angle at vertex 𝐴 is 180βˆ’(30+90)=60∘. In addition, as we have found above that 𝐡𝐷=12𝐴𝐢, it means that 𝐡𝐷=𝐴𝐷.

Hence, △𝐴𝐡𝐷 is isosceles, which means that the two angles formed by each congruent side and the third side are congruent, that is, π‘šβˆ π΄π΅π·=π‘šβˆ π΄=60.∘

This means that the third angle in △𝐴𝐡𝐷 is also of measure 60∘ (since 180βˆ’(60+60)=60∘), and so △𝐴𝐡𝐷 is equilateral. Therefore, we have 𝐴𝐡=𝐡𝐷=𝐴𝐷=24.5.cm

Note that, in the last example, we use the fact that △𝐴𝐡𝐷 is isosceles to prove that π‘šβˆ π΄π΅=60∘. We could also have observed that △𝐡𝐷𝐢 is isosceles, which means that π‘šβˆ π·π΅πΆ=π‘šβˆ πΆ=30.∘

From this, as π‘šβˆ π΄π΅π·=π‘šβˆ π΄π΅πΆβˆ’π‘šβˆ π·π΅πΆ, we deduce that π‘šβˆ π΄π΅π·=90βˆ’30=60.∘

Let us summarize our findings from the last example.

Theorem: The Length of the Smaller Side in a 30Β°-60Β° Right Triangle

In a 30∘-60∘ right triangle, the length of the smaller side (i.e., the side opposite the 30∘ angle) equals half the length of the triangle’s hypotenuse.

In our final example, we will use the properties of medians and their point of concurrence to solve a geometry problem.

Example 5: Finding the Perimeter of a Triangle Using the Medians of the Triangle

Given that 𝐴𝐷=9cm and 𝐸𝐡=𝐴𝐡, find the perimeter of △𝑀𝐷𝐸.

Answer

We are asked to find the perimeter of △𝑀𝐷𝐸. We observe that the lengths of 𝐢𝐷 and 𝐷𝐡 are marked in the diagram to be equal, so 𝐷 is a midpoint. Similarly, 𝐢𝐸=𝐸𝐴, so 𝐸 is a midpoint as well. Hence, 𝐸𝐷 is a line segment joining the midpoints of two sides of △𝐴𝐡𝐢, while 𝐴𝐷 and 𝐡𝐸, which join each vertex of △𝐴𝐡𝐢 to the midpoint of the opposite side, are two medians of △𝐴𝐡𝐢.

Recall that the triangle midsegment theorem states that the segment joining the midpoint of two sides in a triangle is parallel to the third side and is half its length. 𝐸𝐷 is therefore parallel to 𝐴𝐡 and is half its length; that is, half of 12 cm, or 6 cm.

We can now find the length of 𝑀𝐸 and 𝑀𝐷 by recalling that the distance from each vertex of a triangle to the point of concurrence of its medians is two-thirds of the length of the median from this vertex. It follows that the distance from the point of concurrence to the midpoint of one side is one-third of the length of the median from the vertex opposite to this side. Hence, we have 𝑀𝐸=13𝐸𝐡 and 𝑀𝐷=13𝐴𝐷.

It is given that 𝐸𝐡=𝐴𝐡 and it is indicated on the diagram that 𝐴𝐡=12cm. Hence, 𝐸𝐡=12cm. We are also given that 𝐴𝐷=9cm. Substituting these values into the above equations, we get 𝑀𝐸=13Γ—12=4cm and 𝑀𝐷=13Γ—9=3.cm

To find the perimeter of △𝑀𝐷𝐸, we now need to add the lengths of its three sides as follows: Perimetercm=𝐸𝐷+𝑀𝐸+𝑀𝐷=6+4+3=13.

Let us now summarize what we have learned in this explainer.

Key Points

  • The medians of a triangle are the three segments going from each vertex to the midpoint of the opposite side.
  • The three medians of a triangle intersect at a single point: we say that they are concurrent. Their point of intersection is called the point of concurrence of the medians.
  • The distance from each vertex of a triangle to the point of concurrence of its medians is two-thirds of the total length of the median from this vertex. Equivalently, the distance from the point of concurrence of the medians to a vertex is twice that of the distance to the opposite midpoint.
  • In a right triangle, the length of the median drawn from the vertex of the right angle equals half the length of the triangle’s hypotenuse.
  • In a 30∘-60∘ right triangle, the length of the smaller side (i.e., the side opposite the 30∘ angle) equals half the length of the triangle’s hypotenuse.

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