Medians and the Centroid of a Triangle

Centroid of Triangle

Previously, i have written about the Circumcenter and the Incenter, which are two of the four centers of a triangle covered in this blog. The other two centers are the Centroid and the Orthocenter. These four centers are the points in the plane of a triangle with some interesting properties.

Circumcenter is the point equidistant from the three vertices of a triangle. It is also the point where the three perpendicular bisectors of the three sides of the triangle meet.

Circumcenter may lie inside or outside of the triangle depending on what type of triangle it is. To know more about the circumcenter, click here.

Incenter is the point which is equidistant from the three edges. It is also the point of concurrency of the three angle bisectors of the three interior angles of triangle. Unlike Circumcenter, Incenter is always present inside of any triangle.

To know more about the incenter, click here.

Now let’s talk about the Centroid another one of the commonly known triangle centers. Centroid is the point where the three medians of a triangle meet. A median is a line segment connecting a vertex to the midpoint of the opposite edge. A triangle has three medians.

To put it another way, the three medians in any triangle will always intersect at a point(we will prove this fact later in this post). And that point is called the centroid often denoted by capital letter G. Just like incenter, for any type of triangle centroid is always present inside of the triangle.

The centroid can be viewed as similar to how we see the center of a circle. It is the point exactly in the “middle” of the triangle. It is also called the geometric center of the triangle. You might even call it the “balancing point”. Draw a triangle on a thick paper or on a piece of cardboard, and mark the centroid on any one of its two triangular faces. 

Cut off the triangular portion from the cardboard and try to balance this triangular piece on the tip of a pointed object. The triangle will balance at/on the Centroid.

However, this will only be true if the triangular piece of cardboard is of uniform density throughout. If we stick some clay to some part of it, say near any vertex, then the density will no longer remain uniform. The density will be greater in the region near the vertex where clay is applied. And the point of balance will now be shifted towards this higher density region and it will no longer be at/on the centroid.

So technically it is not advisable to think of the centroid as the balance point. However generally it is a good intuitive way of looking at it.

Read : Constructing Centroid in Geogebra

For the remainder of the post, we will look at a proof that Centroid exists for any triangle. But before we begin the proof, we must get familiar with two results from geometry which will be used in the proof.

Result 1 : Diagonals of a parallelogram bisect each other.

A parallelogram is a four sided geometrical shape with opposite sides parallel and equal. One of the properties of a parallelogram is that its diagonals cut each other at their midpoints. This is same as saying that they bisect each other.

Result 2 : Triangle Midpoint Theorem - A line segment joining the midpoints of two adjacent sides of a triangle is parallel to the third side and half the length of the third side.

With the knowledge of these two results let’s prove the centroid.

Consider △ABC. Let P and Q be the midpoints of sides AB and BC. Join BQ and CP. They are the two medians intersecting at a point represented by a letter G.

In order to prove that the three medians meet at a common point, we will have to prove that the third median from the vertex A of the triangle also passes through G. If AR is a segment from A through G, meeting the opposite side at point R, then for AR to be median, R must be the midpoint of side BC.

So in other words if we can prove that R is the midpoint of side BC, then we have proved that the segment AR through G is a median, and hence in turn proved that the three medians are concurrent.

To Prove : R is the midpoint of side BC

Extend segment AR through BC upto point M such that AG = GM. Join MB and MC.

In △ABM, P and G are midpoints of sides AB and AM. PG is the line segment joining the midpoints of sides AB and AM. Therefore, by midpoint theorem of triangles, 

PG ∥ BM

Now points P, G and C are all on the same line segment. So we can say that,

GC ∥ BM              (P-G-C) …….(1)

In △ACM, Q and G are midpoints of sides AC and AM. Therefore, by midpoint theorem of triangles,

    QG ∥ CM

⇒  GB ∥ CM             (Q-G-B) …….(2)

From 1 and 2, the opposite sides of the quadrilateral GBMC are parallel. Therefore, GBMC must be a parallelogram. GM and BC are its diagonals. And from result 1 that we saw earlier, the diagonals of a parallelogram meet at their midpoints, which means R is the midpoint of side BC(& GM).

⇒ R is a mid point of BC

⇒ AR is the median

Hence the three medians in a triangle are concurrent.

(Converse of Ceva’s theorem can also be used to prove the concurrency of Medians of a triangle. In fact, using this theorem is probably the easiest way to prove it. Click here to find out about this theorem and its converse and see its applicability in proving the concurrence of medians.)

So to sum up, a median in a triangle is the line segment drawn from a vertex to the opposite edge. Three medians in a triangle always intersect at a point called the Centroid.