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

The medians of a triangle all pass through the same point. This special point is called the centroid.

A median is a line segment from a vertex to the midpoint of the opposite side.

The Proof Strategy

This proof builds directly on the midsegment theorem:

1. Draw two medians and mark their intersection $G$
2. Cite the midsegment theorem for the segment $MN$ joining two midpoints: $MN \parallel AB$ and $MN = \tfrac{1}{2}AB$
3. Use AA similarity on $\triangle GMN$ and $\triangle GAB$ to prove $G$ divides each median in ratio $2:1$
4. Show the third median must also pass through $G$

The Proof

Step 1: Draw Two Medians

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

• $M$ is the midpoint of $BC$, so median $AM$ goes from vertex $A$ to $M$
• $N$ is the midpoint of $CA$, so median $BN$ 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{AG}{GM} = \frac{BG}{GN} = \frac{CG}{GP} = \frac{2}{1}
    

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

Step 3: The Midsegment $MN$

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

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

• $MN \parallel AB$
• $MN = \tfrac{1}{2} AB$

We take this as given and use the parallel $MN \parallel AB$ as the ingredient for the similarity argument that follows.

Step 4: Similar Triangles

Now observe triangles $GMN$ and $GAB$:

• They share angle $G$ (vertical angles)
• $MN \parallel AB$ creates equal corresponding angles

So $\triangle GMN \sim \triangle GAB$ by AA similarity.

Since $MN = \frac{1}{2}AB$, the ratio of similarity is $1:2$. This means all corresponding sides have ratio $1:2$:

\frac{GM}{GA} = \frac{GN}{GB} = \frac{MN}{AB} = \frac{1}{2}
    

Step 5: The $2:1$ Ratio

If $GA = 2 \cdot GM$, then $G$ divides median $AM$ 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 $BN$:

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

Step 6: The Third Median

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

Using the same midsegment argument with medians $AM$ and $CP$ (and midsegment $MP$), we can show that $G$ divides $CP$ 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!

Conclusion

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

AG = \frac{2}{3} \cdot AM, \quad BG = \frac{2}{3} \cdot BN, \quad CG = \frac{2}{3} \cdot CP
    

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 $G$!