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