common point of triangle medians

Theorem.  The three medians (http://planetmath.org/Median) of a triangle intersect one another in one point, which divides each median in the ratio $2:1$.

Proof.  Let the medians of a triangle  $ABC$  be $AD$, $BE$ and $CF$.  Any median vector is the arithmetic mean of the side vectors emanating from the same vertex.  Using vectors, let us form three ways all beginning from the vertex $A$, the first going simply $2/3$ of the median vector $\overrightarrow{AD}$ ( in the picture):

${\displaystyle \frac{2}{3}}\overrightarrow{AD}={\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{1}{2}}(\overrightarrow{AB}+\overrightarrow{AC})={\displaystyle \frac{1}{3}}(\overrightarrow{AB}+\overrightarrow{AC})$ (1)

The second way goes first the side vector $\overrightarrow{AB}$ and then $2/3$ of the median vector $\overrightarrow{BE}$ (green in the picture):

$\overrightarrow{AB}+{\displaystyle \frac{2}{3}}\overrightarrow{BE}=\overrightarrow{AB}+{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{1}{2}}\left[-\overrightarrow{AB}+(\overrightarrow{AC}-\overrightarrow{AB})\right]={\displaystyle \frac{1}{3}}(\overrightarrow{AB}+\overrightarrow{AC})$ (2)

Similarly, the third way goes first the side vector $\overrightarrow{AC}$ and then $2/3$ of the median vector $\overrightarrow{CF}$ (red in the picture):

$\overrightarrow{AC}+{\displaystyle \frac{2}{3}}\overrightarrow{CF}=\overrightarrow{AC}+{\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{1}{2}}\left[-\overrightarrow{AC}+(\overrightarrow{AB}-\overrightarrow{AC})\right]={\displaystyle \frac{1}{3}}(\overrightarrow{AB}+\overrightarrow{AC})$ (3)

Thus the ways (2) and (3), where one goes from $A$ to another vertex and continues along the corresponding median $2/3$ of its length, lead to the point $M$ which is attained directly along $AD$.  This means that all medians intersect in $M$.  The distance of $M$ from any vertex is $2/3$ of the corresponding median, and so the rest of the median is $1/3$ of its length, i.e. the ratio of the parts of any median is $2:1$.

Title	common point of triangle medians
Canonical name	CommonPointOfTriangleMedians
Date of creation	2013-03-22 17:46:54
Last modified on	2013-03-22 17:46:54
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	8
Author	pahio (2872)
Entry type	Theorem
Classification	msc 51M04
Related topic	MutualPositionsOfVectors
Related topic	ParallelogramPrinciple
Related topic	DifferenceOfVectors
Related topic	TriangleMidSegmentTheorem
Related topic	LengthsOfTriangleMedians