# 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):

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

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

 $\displaystyle\overrightarrow{AC}+\frac{2}{3}\overrightarrow{CF}=% \overrightarrow{AC}+\frac{2}{3}\!\cdot\!\frac{1}{2}\!\left[-\overrightarrow{AC% }+(\overrightarrow{AB}-\overrightarrow{AC})\right]=\frac{1}{3}(\overrightarrow% {AB}+\overrightarrow{AC})$ (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 CommonPointOfTriangleMedians 2013-03-22 17:46:54 2013-03-22 17:46:54 pahio (2872) pahio (2872) 8 pahio (2872) Theorem msc 51M04 MutualPositionsOfVectors ParallelogramPrinciple DifferenceOfVectors TriangleMidSegmentTheorem LengthsOfTriangleMedians