Medians of Triangle Meet at Centroid

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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 $\triangle ABC$.


Corollary

Let $AA'$ be produced beyond $BC$ to $X$, where $A'X = AG$.

Then the straight lines $B'X$ and $CX$ are parallel to $CC'$ and $BB'$ respectively, and $\dfrac 2 3$ of their length.


Proof 1

Medians-meet-at-Centroid.png

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.


Proof 2

Let $\vec a, \vec b, \vec c$ be $\vec{OA}, \vec{OB}, \vec{OC}$ respectively.

Let the midpoint of $BC, AC, AB$ be $\vec d, \vec e, \vec f$ respectively.

Then:

\(\ds \vec d\) \(=\) \(\ds \frac {\vec b + \vec c} 2\)
\(\ds \vec e\) \(=\) \(\ds \frac {\vec a + \vec c} 2\)
\(\ds \vec f\) \(=\) \(\ds \frac {\vec a + \vec b} 2\)

The three medians are $\vec{AD}, \vec{BE}, \vec{CF}$ respectively:

\(\ds \vec {AD}\) \(=\) \(\ds \vec d - \vec a\)
\(\ds \) \(=\) \(\ds \frac {\vec b + \vec c - 2 \vec a} 2\)
\(\ds \vec {BE}\) \(=\) \(\ds \vec e - \vec b\)
\(\ds \) \(=\) \(\ds \frac {\vec a + \vec c - 2 \vec b} 2\)
\(\ds \vec {CF}\) \(=\) \(\ds \vec f - \vec c\)
\(\ds \) \(=\) \(\ds \frac {\vec a + \vec b - 2 \vec c} 2\)

Their equations:

\(\text {(1)}: \quad\) \(\ds \vec {AD}: \ \ \) \(\ds \vec r\) \(=\) \(\ds \vec a + x \paren {\frac {\vec b + \vec c - 2\vec a} 2}\)
\(\text {(2)}: \quad\) \(\ds \vec {BE}: \ \ \) \(\ds \vec r\) \(=\) \(\ds \vec b + y \paren {\frac {\vec a + \vec c - 2\vec b} 2}\)
\(\text {(3)}: \quad\) \(\ds \vec {CF}: \ \ \) \(\ds \vec r\) \(=\) \(\ds \vec c + z \paren {\frac {\vec a + \vec b - 2\vec c} 2}\)

It can be verified that $x = y = z = \dfrac 2 3$ produce the same point:

When $x = \dfrac 2 3$, from $(1)$:

$\vec r = \vec a + \dfrac 2 3 \paren {\dfrac {\vec b + \vec c - 2\vec a} 2} = \dfrac {\vec a + \vec b + \vec c} 3$

When $y = \dfrac 2 3$, from $(2)$:

$\vec r = \vec b + \dfrac 2 3 \paren {\dfrac {\vec a + \vec c - 2\vec b} 2} = \dfrac {\vec a + \vec b + \vec c} 3$

When $z = \dfrac 2 3$, from $(3)$:

$\vec r = \vec c + \dfrac 2 3 \paren {\dfrac {\vec a + \vec b - 2\vec c} 2} = \dfrac {\vec a + \vec b + \vec c} 3$

Therefore, the three medians meet at a single point, namely $\dfrac {\vec a + \vec b + \vec c} 3$.

$\blacksquare$


Also see


Sources

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