Why Medians Meet at One Point

An animated proof showing that all three medians of a triangle intersect at a single point — the centroid — which divides each median in a 2:1 ratio.

Keywords: centroid, medians, triangle centers, 2:1 ratio, similar triangles, midsegment

Prerequisites: midsegment-proof · Difficulty: intermediate

A median of a triangle joins a vertex to the midpoint of the opposite side. The three medians all pass through a single point, the centroid $G$ , which divides each median in a $2 : 1$ ratio from the vertex.

Strategy: draw two medians and mark their intersection $G$ . The midsegment $M N$ joining two midpoints satisfies $M N ∥ A B$ and $M N = \frac{1}{2} A B$ , so $△ GM N \sim △ G A B$ by AA. The ratio $1 : 2$ shows $G$ divides each median $2 : 1$ , and the third median must pass through $G$ as well.

Step 1: Draw Two Medians

We start with triangle $A B C$ and draw two medians. A median connects a vertex to the midpoint of the opposite side.

$M$ is the midpoint of $B C$ , so median $A M$ goes from vertex $A$ to $M$
$N$ is the midpoint of $C A$ , so median $B N$ goes from vertex $B$ to $N$

The two medians intersect at some point — we'll call it $G$ .

Step 2: The Centroid Theorem

The Centroid Theorem says all three medians of a triangle meet at a single point — the centroid $G$ — and $G$ divides each median in ratio $2 : 1$ from the vertex:

\frac{A G}{GM} = \frac{B G}{GN} = \frac{C G}{GP} = \frac{2}{1}

We'll prove it using the midsegment theorem and similar triangles.

Step 3: The Midsegment $M N$

Connect midpoints $M$ and $N$ to form the midsegment $M N$ .

By the midsegment theorem (proven in the previous section):

$M N ∥ A B$
$M N = \frac{1}{2} A B$

We take this as given and use the parallel $M N ∥ A B$ as the ingredient for the similarity argument that follows.

Step 4: Similar Triangles

Now observe triangles $GM N$ and $G A B$ :

They share angle $G$ (vertical angles)
$M N ∥ A B$ creates equal corresponding angles

So $△ GM N \sim △ G A B$ by AA similarity.

Since $M N = \frac{1}{2} A B$ , the ratio of similarity is $1 : 2$ . This means all corresponding sides have ratio $1 : 2$ :

\frac{GM}{G A} = \frac{GN}{GB} = \frac{M N}{A B} = \frac{1}{2}

Step 5: The $2 : 1$ Ratio

If $G A = 2 \cdot GM$ , then $G$ divides median $A M$ so that:

The part from $A$ to $G$ is twice as long as the part from $G$ to $M$
$G$ is located $\frac{2}{3}$ of the way from $A$ to $M$

By the same reasoning applied to median $B N$ :

$G$ divides $B N$ in ratio $2 : 1$ from $B$
$B G = \frac{2}{3} \cdot B N$

Step 6: The Third Median

Now draw the third median from $C$ to $P$ (the midpoint of $A B$ ).

Using the same midsegment argument with medians $A M$ and $C P$ (and midsegment $M P$ ), we can show that $G$ divides $C P$ in ratio $2 : 1$ as well.

Since $G$ is $\frac{2}{3}$ of the way along each median from its vertex, $G$ must lie on all three medians!

Step 7: The Centroid

We've proven that the point $G$ where any two medians intersect must also lie on the third median. Therefore, all three medians meet at a single point.

This point is called the centroid. It divides each median in a $2 : 1$ ratio from the vertex, and it's the triangle's center of mass — the balance point!

All three medians of a triangle meet at the centroid $G$ , which divides each median in a $2 : 1$ ratio from the vertex:

A G = \frac{2}{3} \cdot A M, B G = \frac{2}{3} \cdot B N, C G = \frac{2}{3} \cdot C P

The centroid is the center of mass of the triangle — if you made the triangle from uniform material, it would balance on a pin at $G$ ! ∎