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# common point of triangle medians

Theorem. The three medians 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 $\overrightarrow{AD}$ (blue 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 $2\!:\!1$.

## Mathematics Subject Classification

51M04*no label found*

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