Centroid - concurrency of the medians

The centroid is the point where the three medians of the triangle meet. We prove that the medians are indeed concurrent and that the common point divides each median in ratio $2:1$.

Theorem — Concurrency of the medians

The three medians of a triangle meet at a single point $G$ (centroid), which divides each median in ratio $\dfrac{2}{1}$ starting from the vertex.

Proof — The medians are concurrent

1. Let $M$ be the midpoint of $AC$ and $N$ the midpoint of $BC$. The medians $BM$ and $AN$ intersect at a point $G$.
2. By the theorem on the intersection of two medians, $G$ divides $AN$ so that $AG\cong 2\,GN$ and $BG\cong 2\,GM$.
3. Now join $C$ with $G$ and extend until it meets $AB$ at $K$.
4. If from $C$ we drew the median (towards the midpoint of $AB$), it would, by the same theorem, intersect $AN$ at a point $S$ such that $AS = 2\,SN$. But $G$ is the only point of $AN$ with this property; hence $S$ coincides with $G$.
5. So $CK$ too passes through $G$, and $K$ is the midpoint of $AB$: all three medians are concurrent at $G$.

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Topics: Euclidean circle
Concepts: Centroid · Median
Skills: Proving · Synthetic geometry