# Lesson Explainer: Medians of Triangles Mathematics

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 and then, from the second equation (by multiplying both sides by 3), we get and by substituting with into the first equation, we get which gives .

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 , is the point of concurrency of its medians. If is a median, then .

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 , with as their point of concurrence. is the median that joins vertex with ; hence, is the midpoint of side .

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

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

Hence, .

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 , given that .

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

Recall that the length of , that is, the distance from the vertex to the point of concurrence, is two-thirds that of the median :

Thus, the length of 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 , and . Find .

In triangle , is the point of concurrence of its medians. is a vertex and is the middle of the opposite side, . 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 , we have and

As we see that

It is said in the question that and

Hence, we have

Dividing both sides of this equation by 2 gives

Adding 7 to both sides gives

And, finally, dividing both sides by 5, we find that

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 is a rectangle and and are congruent right triangles.

In , is a median. In rectangle , 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 and are congruent), we have , so , that is, .

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 triangleβ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 - 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 and .

From the diagram, we observe that is a right triangle at , and as is the midpoint of , is the median of 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 triangleβs hypotenuse. Hence, we have

We are asked as well to find the length of . The measure of the angle at vertex is given in the diagram: it is . As the angle at is a right angle, it means that the measure of the angle at vertex is . In addition, as we have found above that , it means that .

Hence, is isosceles, which means that the two angles formed by each congruent side and the third side are congruent, that is,

This means that the third angle in is also of measure (since ), and so is equilateral. Therefore, we have

Note that, in the last example, we use the fact that is isosceles to prove that . We could also have observed that is isosceles, which means that

From this, as , we deduce that

Let us summarize our findings from the last example.

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

In a - right triangle, the length of the smaller side (i.e., the side opposite the angle) equals half the length of the triangleβ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 and , find the perimeter of .

We are asked to find the perimeter of . We observe that the lengths of and are marked in the diagram to be equal, so is a midpoint. Similarly, , so is a midpoint as well. Hence, is a line segment joining the midpoints of two sides of , while and , which join each vertex of to the midpoint of the opposite side, are two medians of .

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. is therefore parallel to and is half its length; that is, half of 12 cm, or 6 cm.

We can now find the length of and 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 and

It is given that and it is indicated on the diagram that . Hence, . We are also given that . Substituting these values into the above equations, we get and

To find the perimeter of , we now need to add the lengths of its three sides as follows:

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 triangleβs hypotenuse.
• In a - right triangle, the length of the smaller side (i.e., the side opposite the angle) equals half the length of the triangleβs hypotenuse.