# Medians of Triangle Meet at Centroid/Proof 1

## Theorem

Let $\triangle ABC$ be a triangle.

Then the medians of $\triangle ABC$ meet at a single point.

This point is called the centroid of $\triangle ABC$.

## Proof

Let $A'$ be the midpoint of $BC$.

Let $B'$ be the midpoint of $AC$.

Let $C'$ be the midpoint of $AB$.

Hence $AA'$, $BB'$ and $CC'$ are the medians of $\triangle ABC$.

Let $AA'$ and $BB'$ intersect at $G$.

Hence $A'B'$ is a midline of $\triangle ABC$.

By the Midline Theorem, $A'B'$ is parallel to $AB$ and half the length of $AB$.

We have that $\triangle AGB$ and $\triangle A'GB'$ are similar.

Hence:

\(\ds A'G\) | \(=\) | \(\ds \dfrac {AG} 2\) | ||||||||||||

\(\ds B'G\) | \(=\) | \(\ds \dfrac {BG} 2\) |

Hence $BB'$ meets $AA'$ $\dfrac 1 3$ of the distance along $AA'$ from $A'$.

That is:

- $A'G = \dfrac {AA'} 3$

Similarly, *mutatis mutandis*, it can be shown that $CC'$ also meets $AA'$ $\dfrac 1 3$ of the distance along $AA'$ from $A'$.

Hence we have shown that $BB'$ and $CC'$ both meet at the same point on $AA'$

That is, the medians of $\triangle ABC$ meet at a single point.

## Sources

- 1953: L. Harwood Clarke:
*A Note Book in Pure Mathematics*... (previous) ... (next): $\text {IV}$. Pure Geometry: Plane Geometry: The centre of gravity